題目大意：給定 r * c 的迷宮，還有一個整數 k 。迷宮中“.”表示可以走，“#”表示牆，當時間為k的倍數時，這些牆會消失。求從起點“Y”到終點“G”的最短時間。（人不能呆在一點不動）。
[cpp]
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define N 105
using namespace std;
char map[N][N];
int step[N][N][10];//多加一維，記錄(time%k)
int r,c,k,x_s,y_s,x_e,y_e;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
struct node
{
int x,y,mod;
};
int BFS()
{
int i;
queue<node>q;
node now,next;
now.x=x_s;
now.y=y_s;
now.mod=0;
memset(step,-1,sizeof(step));
step[now.x][now.y][now.mod]=0;
q.push(now);
while(!q.empty()){
now=q.front();
q.pop();
if(now.x==x_e && now.y==y_e) return step[now.x][now.y][now.mod];
for(i=0;i<4;i++){
next.x=now.x+dir[i][0];
next.y=now.y+dir[i][1];
next.mod=(now.mod+1)%k;
if(next.x<0 || next.x>=r || next.y<0 ||next.y>=c) continue;
if(step[next.x][next.y][next.mod]!=-1) continue;
if(map[next.x][next.y]=='#' && next.mod!=0) continue;
step[next.x][next.y][next.mod]=step[now.x][now.y][now.mod]+1;
q.push(next);
}
}
return -1;
}
int main()
{
int T,i,j,ans;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&r,&c,&k);
for(i=0;i<r;i++){
scanf("%s",map[i]);
for(j=0;map[i][j];j++){
if(map[i][j]=='Y'){
x_s=i;y_s=j;map[i][j]='.';
}
if(map[i][j]=='G'){
x_e=i;y_e=j;map[i][j]='.';
}
}
}
ans=BFS();
if(ans==-1) printf("Please give me another chance!\n");
else printf("%d\n",ans);
}
return 0;
}
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define N 105
using namespace std;
char map[N][N];
int step[N][N][10];//多加一維，記錄(time%k)
int r,c,k,x_s,y_s,x_e,y_e;
int dir[4][2]={0,1,0,-1,1,0,-1,0};
struct node
{
int x,y,mod;
};
int BFS()
{
int i;
queue<node>q;
node now,next;
now.x=x_s;
now.y=y_s;
now.mod=0;
memset(step,-1,sizeof(step));
step[now.x][now.y][now.mod]=0;
q.push(now);
while(!q.empty()){
now=q.front();
q.pop();
if(now.x==x_e && now.y==y_e) return step[now.x][now.y][now.mod];
for(i=0;i<4;i++){
next.x=now.x+dir[i][0];
next.y=now.y+dir[i][1];
next.mod=(now.mod+1)%k;
if(next.x<0 || next.x>=r || next.y<0 ||next.y>=c) continue;
if(step[next.x][next.y][next.mod]!=-1) continue;
if(map[next.x][next.y]=='#' && next.mod!=0) continue;
step[next.x][next.y][next.mod]=step[now.x][now.y][now.mod]+1;
q.push(next);
}
}
return -1;
}
int main()
{
int T,i,j,ans;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&r,&c,&k);
for(i=0;i<r;i++){
scanf("%s",map[i]);
for(j=0;map[i][j];j++){
if(map[i][j]=='Y'){
x_s=i;y_s=j;map[i][j]='.';
}
if(map[i][j]=='G'){
x_e=i;y_e=j;map[i][j]='.';
}
}
}
ans=BFS();
if(ans==-1) printf("Please give me another chance!\n");
else printf("%d\n",ans);
}
return 0;
}
