## A - Red and Black

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0


Sample Output

45
59
6
13


### CODE

#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <vector>
#include <set>
#define PI acos(-1)
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define debug freopen("in.txt","r",stdin); freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int N = 210;
const int MOD = 1e9;
const double eps = 1e-5;
const int INF = 0x3f3f3f;
struct node{
int x,y,step;
};
int n,m;
char Map[N][N];
bool vis[N][N];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int bfs(int sx,int sy){
int cnt=0;
node s={sx,sy,0};
queue<node> q;
q.push(s);
vis[sx][sy]=1;
while(!q.empty()){
node fr=q.front();q.pop();
cnt++;
for(int i=0;i<4;i++){
node next={fr.x+dir[i][0],fr.y+dir[i][1],fr.step+1};
if(next.x<1||next.x>n||next.y>m||next.y<1) continue;
if(vis[next.x][next.y]||Map[next.x][next.y]=='#') continue;
vis[next.x][next.y]=1;
q.push(next);
}
}
return cnt;
}
int main(){
ios;
while(cin>>m>>n){
if(m==0&n==0) break;
memset(vis,0,sizeof vis);
int sx,sy;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>Map[i][j];
if(Map[i][j]=='@'){
sx=i;
sy=j;
}
}
}
int ans=bfs(sx,sy);
if(ans!=-1) cout<<ans<<'\n';
else cout<<"oop!"<<'\n';
}
return 0;
}


## B - Rescue

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel’s friends want to save Angel. Their task is: approach Angel. We assume that “approach Angel” is to get to the position where Angel stays. When there’s a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. “.” stands for road, “a” stands for Angel, and “r” stands for each of Angel’s friend.

Process to the end of the file.

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing “Poor ANGEL has to stay in the prison all his life.”

Sample Input

7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........


Sample Output

13


### 分析

#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <vector>
#include <set>
#define PI acos(-1)
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define debug freopen("in.txt","r",stdin); freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int N = 210;
const int MOD = 1e9;
const double eps = 1e-5;
const int INF = 0x3f3f3f;
struct node{
int x,y,time;
bool operator < (const node &o) const {
return o.time<time;
}
};
int n,m;
char Map[N][N];
bool vis[N][N];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int bfs(int sx,int sy){
node s={sx,sy,0};
priority_queue<node> q;
q.push(s);
vis[sx][sy]=1;
while(!q.empty()){
node fr=q.top();q.pop();
for(int i=0;i<4;i++){
node next={fr.x+dir[i][0],fr.y+dir[i][1],fr.time+1};
if(next.x<1||next.x>n||next.y>m||next.y<1) continue;
if(vis[next.x][next.y]||Map[next.x][next.y]=='#') continue;
if(Map[next.x][next.y]=='x') next.time++;
if(Map[next.x][next.y]=='a'){
return next.time;
}
vis[next.x][next.y]=1;
q.push(next);
}
}
return INF;
}
int main(){
ios;
while(cin>>n>>m){
memset(vis,0,sizeof vis);
int sx,sy;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>Map[i][j];
if(Map[i][j]=='r'){
sx=i;
sy=j;
}
}
}
int ans=bfs(sx,sy);
if(ans!=INF) cout<<ans<<'\n';
else cout<<"Poor ANGEL has to stay in the prison all his life."<<'\n';
}
return 0;
}


## C - Battle City

Many of us had played the game “Battle city” in our childhood, and some people (like me) even often play it on computer now.

What we are discussing is a simple edition of this game. Given a map that consists of empty spaces, rivers, steel walls and brick walls only. Your task is to get a bonus as soon as possible suppose that no enemies will disturb you (See the following picture).

Your tank can’t move through rivers or walls, but it can destroy brick walls by shooting. A brick wall will be turned into empty spaces when you hit it, however, if your shot hit a steel wall, there will be no damage to the wall. In each of your turns, you can choose to move to a neighboring (4 directions, not 8) empty space, or shoot in one of the four directions without a move. The shot will go ahead in that direction, until it go out of the map or hit a wall. If the shot hits a brick wall, the wall will disappear (i.e., in this turn). Well, given the description of a map, the positions of your tank and the target, how many turns will you take at least to arrive there?

Input

The input consists of several test cases. The first line of each test case contains two integers M and N (2 <= M, N <= 300). Each of the following M lines contains N uppercase letters, each of which is one of ‘Y’ (you), ‘T’ (target), ‘S’ (steel wall), ‘B’ (brick wall), ‘R’ (river) and ‘E’ (empty space). Both ‘Y’ and ‘T’ appear only once. A test case of M = N = 0 indicates the end of input, and should not be processed.

Output

For each test case, please output the turns you take at least in a separate line. If you can’t arrive at the target, output “-1” instead.

Sample Input

3 4
YBEB
EERE
SSTE
0 0


Sample Output

8


### CODE

#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <vector>
#include <set>
#define PI acos(-1)
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define debug freopen("in.txt","r",stdin); freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int N = 410;
const int MOD = 1e9;
const double eps = 1e-5;
const int INF = 0x3f3f3f;
struct node{
int x,y,time;
bool operator < (const node &o) const {
return o.time<time;
}
};
int n,m;
char Map[N][N];
bool vis[N][N];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int bfs(int sx,int sy){
node s={sx,sy,0};
priority_queue<node> q;
q.push(s);
vis[sx][sy]=1;
while(!q.empty()){
node fr=q.top();q.pop();
for(int i=0;i<4;i++){
node next={fr.x+dir[i][0],fr.y+dir[i][1],fr.time+1};
if(next.x<1||next.x>n||next.y>m||next.y<1) continue;
if(vis[next.x][next.y]||Map[next.x][next.y]=='S'||Map[next.x][next.y]=='R') continue;
if(Map[next.x][next.y]=='B') next.time++;
if(Map[next.x][next.y]=='T'){
return next.time;
}
vis[next.x][next.y]=1;
q.push(next);
}
}
return -1;
}
int main(){
ios;
while(cin>>n>>m){
if(n==0&&m==0) break;
memset(vis,0,sizeof vis);
int sx,sy;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>Map[i][j];
if(Map[i][j]=='Y'){
sx=i;
sy=j;
}
}
}
int ans=bfs(sx,sy);
cout<<ans<<'\n';
}
return 0;
}


## Dungeon Master

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a ‘#’ and empty cells are represented by a ‘.’. Your starting position is indicated by ‘S’ and the exit by the letter ‘E’. There’s a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

​ Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line

​ Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0


Sample Output

Escaped in 11 minute(s).
Trapped!


### CODE

#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <vector>
#include <set>
#define PI acos(-1)
#define ios ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define debug freopen("in.txt","r",stdin); freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int N = 40;
const int MOD = 1e9;
const double eps = 1e-5;
const int INF = 0x3f3f3f;
struct node{
int x,y,z,time;
};
int l,n,m;
char Map[N][N][N];
bool vis[N][N][N];
int dir[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
bool check(node next){
if(next.x<1||next.x>n||next.y>m||next.y<1||next.z<1||next.z>l) return 0;
else return 1;
}
int bfs(int sx,int sy,int sz){
node s={sx,sy,sz,0};
queue<node> q;
q.push(s);
vis[sz][sx][sy]=1;
while(!q.empty()){
node fr=q.front();q.pop();
for(int i=0;i<6;i++){
node next={fr.x+dir[i][0],fr.y+dir[i][1],fr.z+dir[i][2],fr.time+1};
if(!check(next)) continue;
if(vis[next.z][next.x][next.y]||Map[next.z][next.x][next.y]=='#') continue;
if(Map[next.z][next.x][next.y]=='E'){
return next.time;
}
vis[next.z][next.x][next.y]=1;
q.push(next);
}
}
return -1;
}
int main(){
//	ios;
while(cin>>l>>n>>m){
if(l==0&&n==0&&m==0) break;
memset(vis,0,sizeof vis);
int sx,sy,sz;
for(int k=1;k<=l;k++){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>Map[k][i][j];
if(Map[k][i][j]=='S'){
sx=i;
sy=j;
sz=k;
}
}
}
}
int ans=bfs(sx,sy,sz);
if(ans!=-1) cout<<"Escaped in "<<ans<<" minute(s)."<<'\n';
else cout<<"Trapped!"<<'\n';
}
return 0;
}


## Catch That Cow

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17


Sample Output

4


### CODE

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
int vis[1000001];
struct node{
int wei;
int time;
};
int main()
{
int n,k;
cin>>n>>k;
node start={n,0};
vis[n]=1;
queue<node>q;
q.push(start);
while(!q.empty()){
node tmp=q.front(); q.pop();
if(tmp.wei==k){
cout<<tmp.time<<'\n';
break;
}
for(int i=0;i<3;i++){
node nx;
if(i==0) nx={tmp.wei+1,tmp.time+1};
else if(i==1) nx={tmp.wei-1,tmp.time+1};
else nx={tmp.wei*2,tmp.time+1};
if(nx.wei<0||nx.wei>100000||vis[nx.wei]) continue;
vis[nx.wei]=1;
q.push(nx);
}
}
return 0;
}


## Number Transformation

In this problem, you are given an integer number s. You can transform any integer number A to another integer number B by adding x to A. This x is an integer number which is a prime factor of A (please note that 1 and A are not being considered as a factor of A). Now, your task is to find the minimum number of transformations required to transform s to another integer number t.

Input

Input starts with an integer T (****≤ 500), denoting the number of test cases.

Each case contains two integers: s (1 ≤ s ≤ 100) and t (1 ≤ t ≤ 1000).

Output

For each case, print the case number and the minimum number of transformations needed. If it’s impossible, then print -1.

Sample Input

2

6 12

6 13


Sample Output

Case 1: 2

Case 2: -1


### CODE

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;
int T,t,s;
const int MAX=1e3+10;
const int INF=0x3f3f3f;
int vis[MAX];
vector<int> vt[MAX];

struct node{
int x,step;
};
bool check(int n){
if(n==2||n==3) return 1;
if(n==1||n%6!=1&&n%6!=5) return 0;
for(int i=5;i*i<=n;i++){
if(n%i==0||n%(i+2)==0) return 0;
}
return 1;
}
void prime(){
for(int i=2;i<MAX;i++){
for(int j=2;j<i;j++){
if(i%j==0&&check(j)) vt[i].push_back(j);
}
}
}
int bfs(){
queue<node> q;
vis[s]=1;
node start;
start.x=s; start.step=0;
q.push(start);
while(!q.empty()){
node tmp=q.front(); q.pop();
if(tmp.x==t) return tmp.step;
int len=vt[tmp.x].size();
for(int i=0;i<len;i++){
node nx;
nx.x=tmp.x+vt[tmp.x][i];
nx.step=tmp.step+1;
if(nx.x>t||vis[nx.x]) continue;
vis[nx.x]=1;
q.push(nx);
}
}
return -1;
}
int main()
{
prime();
cin>>T;
int kase=0;
while(T--){
memset(vis,0,sizeof(vis));
scanf("%d%d",&s,&t);
int ans=INF;
ans=bfs();
printf("Case %d: %d\n",++kase,ans);
}
return 0;
}


## G - Knight Moves

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the “difficult” part.

Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.

Input

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

Output

For each test case, print one line saying “To get from xx to yy takes n knight moves.”.

Sample Input

e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6


Sample Output

To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.


### CODE

#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
char ch1[3],ch2[3];
int vis[9][9],ans;
int dir[8][2]={{2,1},{-2,1},{2,-1},{-2,-1},{1,2},{1,-2},{-1,2},{-1,-2}};
struct node{
int row,col,step;
};
void bfs(){
queue<node> q;
node s;
s.row=ch1[1]-'1';
s.col=ch1[0]-'a';
s.step=0;
vis[s.row][s.col]=1;
q.push(s);
while(!q.empty()){
node tmp=q.front(); q.pop();
if(tmp.row==(ch2[1]-'1')&&tmp.col==(ch2[0]-'a')){
if(tmp.step<ans) ans=tmp.step;
return;
}
for(int i=0;i<8;i++){
node nx;
nx.row=tmp.row+dir[i][0];
nx.col=tmp.col+dir[i][1];
if(!vis[nx.row][nx.col]&&nx.row>=0&&nx.row<8&&nx.col>=0&&nx.col<8){
vis[nx.row][nx.col]=1;
nx.step=tmp.step+1;
q.push(nx);
}
}
}
}
int main()
{
while(cin>>ch1>>ch2){
ans=1e9;
memset(vis,0,sizeof(vis));
bfs();
cout<<"To get from "<<ch1<<" to "<<ch2<<" takes "<<ans<<" knight moves."<<'\n';
}
return 0;
}


## 噩梦

#### 数据范围

1<n,m<8001<n,m<800

#### 输入样例：

3
5 6
XXXXXX
XZ..ZX
XXXXXX
M.G...
......
5 6
XXXXXX
XZZ..X
XXXXXX
M.....
..G...
10 10
..........
..X.......
..M.X...X.
X.........
.X..X.X.X.
.........X
..XX....X.
X....G...X
...ZX.X...
...Z..X..X


#### 输出样例：

1
1
-1


### CODE

#include<bits/stdc++.h>
using namespace std;
const int N=810;
typedef pair<int,int> pii;
int n,m;
pii boy,girl,ghost[2];
int vis[N][N],dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
char g[N][N];
bool check(int x,int y,int k){
if(x<1||x>n||y<1||y>m||g[x][y]=='X') return false;
for(int i=0;i<2;i++){
if(abs(x-ghost[i].first)+abs(y-ghost[i].second)<=2*k) return false;
}
return true;
}
int bfs(){
queue<pii> qb,qg;
qb.push(boy); qg.push(girl);
int step=0;
while(!qb.empty()||!qg.empty()){
step++;
for(int k=0;k<3;k++){
for(int j=0,len=qb.size();j<len;j++){
int x=qb.front().first;
int y=qb.front().second;
qb.pop();
if(!check(x,y,step)) continue;
for(int i=0;i<4;i++){
int nx=x+dx[i];
int ny=y+dy[i];
if(!check(nx,ny,step)) continue;
if(vis[nx][ny]==2) return step;
if(!vis[nx][ny]){
vis[nx][ny]=1;
qb.push({nx,ny});
}
}
}
}
for(int k=0;k<1;k++){
for(int j=0,len=qg.size();j<len;j++){
int x=qg.front().first;
int y=qg.front().second;
qg.pop();
if(!check(x,y,step)) continue;
for(int i=0;i<4;i++){
int nx=x+dx[i];
int ny=y+dy[i];
if(!check(nx,ny,step)) continue;
if(vis[nx][ny]==1) return step;
if(!vis[nx][ny]){
vis[nx][ny]=2;
qg.push({nx,ny});
}

}
}
}
}
return -1;
}
int main()
{
int t;
cin>>t;
while(t--){
memset(vis,0,sizeof vis);
cin>>n>>m;
int cnt=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
cin>>g[i][j];
if(g[i][j]=='M') boy={i,j};
if(g[i][j]=='G') girl={i,j};
if(g[i][j]=='Z') ghost[cnt++]={i,j};
}
}
cout<<bfs()<<'\n';
}
return 0;
}