OpenJudge noi 1159Maze(poj 2157)
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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 mazeInput
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 OutputYES
NO
bfs:
这个题,可以单独开一个结构体,记录门的位置。在输入过程中,我们先把门处理一下,放入结构体中,然后将图遍历一遍,再处理每个门的钥匙个数。在bfs的过程中,如果我们找到了门,但是没有找够钥匙,我们就把这个门的vis记录为1,如果可以找够钥匙,我们就再把这个门的坐标放到队列中,让它再继续拓展;若我们在找够钥匙之后,发现门还没有没访问过,我们就直接把门变为一条路,直到我们找到目标位置。
最重要的一点:初始化的时候队列别忘了清空!!!mdzz,这是第二次犯这种问题了,谨记!!!
代码如下:
#include<iostream>#include<cstdio>#include<cstdlib>#include<queue>#include<cstring>using namespace std;const int maxn=550;int dx[]={0,1,0,-1,0};int dy[]={0,0,1,0,-1};struct dqs{ int x,y;};struct dqm{ int x,y,key; bool vis;}door[maxn];char map[maxn][maxn];bool use[maxn][maxn];int sx,sy; int n,m;queue<dqs>q;void open(int z){ if(door[z].vis==1) { dqs nxt; nxt.x=door[z].x; nxt.y=door[z].y; q.push(nxt); } else map[door[z].x][door[z].y]='.';}bool check(int x,int y){ if(x>=1&&x<=n&&y>=1&&y<=m&&map[x][y]!='X'&&use[x][y]==0) return true; return false;}void bfs(){ while(!q.empty()) q.pop(); dqs begin; begin.x=sx,begin.y=sy; q.push(begin); use[sx][sy]=1; while(!q.empty()) { dqs head=q.front(); q.pop(); dqs now; for(int i=1;i<=4;i++) { now.x=head.x+dx[i]; now.y=head.y+dy[i]; if(check(now.x,now.y)) { if(map[now.x][now.y]=='G') { printf("YES\n"); return; } else { use[now.x][now.y]=1; if(map[now.x][now.y]>='a'&&map[now.x][now.y]<='e') { q.push(now); int ss=map[now.x][now.y]-'a'+1; door[ss].key--; if(door[ss].key==0) open(ss); } if(map[now.x][now.y]>='A'&&map[now.x][now.y]<='E') { int ss=map[now.x][now.y]-'0'-16; door[ss].vis=1; } if(map[now.x][now.y]=='.') q.push(now); } } } } printf("NO\n"); return;}int main(){ string a; while(1) { scanf("%d%d",&n,&m); if(n==0&&m==0) break; memset(map,0,sizeof(map)); memset(use,0,sizeof(use)); memset(door,0,sizeof(door)); for(int i=1;i<=n;i++) { cin>>a; for(int j=0;j<m;j++) { map[i][j+1]=a[j]; if(map[i][j+1]>='A'&&map[i][j+1]<='E') { int ss=map[i][j+1]-'0'-16; door[ss].x=i; door[ss].y=j+1; door[ss].vis=0; door[ss].key=0; } } } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { if(map[i][j]=='S') sx=i,sy=j; if(map[i][j]>='a'&&map[i][j]<='e') door[map[i][j]-'0'-48].key++; } bfs(); } return 0;}
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