POJ2157Maze[DFS !]

Posted 2020-08-12 Candy?

Maze

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3818		Accepted: 1208

Description

Acm, a treasure-explorer, is exploring again. This time he is in a special maze, in which there are some doors (at most 5 doors, represented by ‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘ respectively). In order to find the treasure, Acm may need to open doors. However, to open a door he needs to find all the door‘s keys (at least one) in the maze first. For example, if there are 3 keys of Door A, to open the door he should find all the 3 keys first (that‘s three ‘a‘s which denote the keys of ‘A‘ in the maze). Now make a program to tell Acm whether he can find the treasure or not. Notice that Acm can only go up, down, left and right in the maze. 

Input

The input consists of multiple test cases. The first line of each test case contains two integers M and N (1 < N, M < 20), which denote the size of the maze. The next M lines give the maze layout, with each line containing N characters. A character is one of the following: ‘X‘ (a block of wall, which the explorer cannot enter), ‘.‘ (an empty block), ‘S‘ (the start point of Acm), ‘G‘ (the position of treasure), ‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘ (the doors), ‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘ (the keys of the doors). The input is terminated with two 0‘s. This test case should not be processed. 

Output

For each test case, in one line output "YES" if Acm can find the treasure, or "NO" otherwise. 

Sample Input

4 4 
S.X. 
a.X. 
..XG 
.... 
3 4 
S.Xa 
.aXB 
b.AG 
0 0

Sample Output

YES 
NO

Source

POJ Monthly,Wang Yijie

---

题意：迷宫，有门有钥匙

---

1.多个G，不能简单记录终点

2.有的门没钥匙，就是不能进

不需要回溯，一边搜下去，遇到钥匙判断一下这个钥匙凑齐了没有，如果凑齐了则增加从这个钥匙能打开的门开始搜索(用mark表示这个门有没有到达过)

 

//
//  main.cpp
//  poj2157
//
//  Created by Candy on 10/1/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
using namespace std;
const int N=21,V=1e6+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x;
}
int n,m,sx,sy,flag=0;
char a[N][N];
struct doors{
    int k,x,y;
}door[N*N];
int cnt=0;
int vis[N][N],g[N][N],dx[4]={-1,1,0,0},dy[4]={0,0,1,-1};
struct keys{
    int has,tot;
}key[N];
int mark[N][N];
void dfs(int x,int y){//printf("dfs %d %d\n",x,y);
    vis[x][y]=1;
    if(a[x][y]==‘G‘){flag=1;return;}
    if(flag) return;
    for(int i=0;i<4;i++){
        int nx=x+dx[i],ny=y+dy[i];
        if(vis[nx][ny]) continue;
        if(nx<1||nx>n||ny<1||ny>m) continue;
        if(g[nx][ny]<=5&&g[nx][ny]>=1){
            int num=g[nx][ny];
            mark[nx][ny]=1;
            if(key[num].has<key[num].tot||(!key[num].tot)) continue;
        }
        if(g[nx][ny]>5) {
            int num=g[nx][ny]-5;
            key[num].has++;
            if(key[num].has==key[num].tot&&key[num].tot){//no key cannot in
                for(int j=1;j<=cnt;j++)
                    if(door[j].k==num&&mark[door[j].x][door[j].y]){
                        dfs(door[j].x,door[j].y);
                    }
            }
        }
        dfs(nx,ny);
    }
    //printf("end %d %d\n",x,y);
}
int main(int argc, const char * argv[]) {
    while(scanf("%d%d",&n,&m)&&n){
        memset(g,0,sizeof(g));
        memset(vis,0,sizeof(vis));
        memset(door,0,sizeof(door));
        memset(key,0,sizeof(key));
        memset(mark,0,sizeof(mark));
        cnt=0;
        for(int i=1;i<=n;i++){
            flag=0;
            scanf("%s",a[i]+1);
            for(int j=1;j<=m;j++){
                if(a[i][j]==‘S‘) sx=i,sy=j;
                else if(a[i][j]==‘X‘) vis[i][j]=1;
                else if(a[i][j]>=‘A‘&&a[i][j]<=‘E‘){
                    g[i][j]=a[i][j]-‘A‘+1;//1...5
                    door[++cnt].x=i;door[cnt].y=j;door[cnt].k=g[i][j];
                }else if(a[i][j]>=‘a‘&&a[i][j]<=‘e‘){
                    key[a[i][j]-‘a‘+1].tot++;
                    g[i][j]=a[i][j]-‘a‘+6;//6...10
                }
            }
        }
        dfs(sx,sy);
        if(flag) puts("YES");
        else puts("NO");
        //printf("%d %d %d",key[2].has,key[2].tot,door[2].k);
    }
    
    return 0;
}