poj2157Maze(bfs,有門，需要先找鑰匙)

Maze

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3245		Accepted: 1022

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

dfs，若找到G，標記之，循環bfs，根據跳出條件執行，鑰匙增加就循環，不增加就跳出。

再根據標記，輸出結果。dfs，bfs應該都行吧。

#include <iostream>
#include <cstring>
#include <cctype>
using namespace std;
bool vis[20][20];
int dir[4][2]={-1,0,1,0,0,-1,0,1};
char mat[25][25];
int r,c,sx,sy,ex,ey,flag,ok,p;
int totalKey[5], haveKey[5];
void dfs(int x, int y){
    if(x==ex && y==ey)  ok=1;
    if(mat[x][y]=='X' || vis[x][y]) return;
    if(islower(mat[x][y])) {haveKey[mat[x][y]-'a']++, mat[x][y]='.',flag++;}
    if(mat[x][y]=='A' || mat[x][y]=='B' || mat[x][y]=='C' || mat[x][y]=='D' || mat[x][y]=='E'){
        if(haveKey[mat[x][y]-'A']==totalKey[mat[x][y]-'A'] && haveKey[mat[x][y]-'A']!=0) mat[x][y]='.';  //&& 寫成 || 找了半天錯
        else return;
    }
    vis[x][y] = true;
    for(int i=0; i<4; i++){
        int tx = x + dir[i][0];
        int ty = y + dir[i][1];
        if(tx<0 || tx>r-1 || ty<0 || ty>c-1) continue;
        dfs(tx,ty);
    }
}

int main()
{
    while(cin>>r>>c &&r+c){
        flag=ok=0;
        memset(vis,false,sizeof(vis));
        memset(mat,'\0',sizeof(mat));
        memset(haveKey,0,sizeof(haveKey));
        memset(totalKey,0,sizeof(totalKey));
        for(int i=0; i<r; i++){
            for(int j=0; j<c; j++){
                cin>>mat[i][j];
                if(islower(mat[i][j])) totalKey[mat[i][j]-'a']++;
                if(mat[i][j]=='S') sx = i, sy = j, mat[i][j]='.';
                if(mat[i][j]=='G') ex = i, ey = j;                     //一開始ey寫成sy，找了半天錯
            }
        }

        dfs(sx,sy);p=flag;
        while(true){
            memset(vis,false,sizeof(vis));
            dfs(sx,sy);
            if(flag==p) break;
            else p=flag;
        }
        
//        int m = 405;                       //棋盤是20*20的，所以最多搜索400次，其實肯定不可能這麼多，懶得推證了
//        while(m--){
//            dfs(sx,sy);
//            memset(vis,false,sizeof(vis));
//        }

        if(ok) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;

    }
    return 0;
}