题意: 公主被关在 a位置 她的朋友在r位置
路上x位置有恶魔
遇上恶魔花费2 时间 否在时间花费 1 时间
问 最短多少时间 找到公主
思路: bfs+ 优先队列(时间短的先出列)
#include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<iostream> #include<algorithm> using namespace std; char mat[300][300]; int vis[300][300]; struct Node { int x,y,coun; friend bool operator < (Node a,Node b){ return a.coun>b.coun; } //小的先出列 }; int sx,sy,ex,ey; int op[4][2]={1,0,-1,0,0,1,0,-1}; int n,m; int ans; bool fit(Node a) { if(0<=a.x&&a.x<n&&0<=a.y&&a.y<m&&vis[a.x][a.y]==0&&mat[a.x][a.y]!='#') return true; return false; } void bfs() { priority_queue<Node> q; Node f; f.x=sx; f.y=sy; f.coun=0; q.push(f); while(!q.empty()) { Node now=q.top(); q.pop(); vis[now.x][now.y]=1; if(mat[now.x][now.y]=='r') {ans=now.coun;break;} for(int i=0;i<4;i++) { Node next; next.x=now.x+op[i][0]; next.y=now.y+op[i][1]; if(fit(next)) { if(mat[next.x][next.y]=='.'||mat[next.x][next.y]=='r') next.coun=now.coun+1; else if(mat[next.x][next.y]=='x') next.coun=now.coun+2; q.push(next); } } } } int main() { int i,j,k; while(scanf("%d%d",&n,&m)!=EOF) { ans=-1; memset(vis,0,sizeof(vis)); for(i=0;i<n;i++) { scanf("%s",mat[i]); for(j=0;j<m;j++) { if(mat[i][j]=='a') {sx=i;sy=j;} } } bfs(); if(ans>-1) printf("%d ",ans); else printf("Poor ANGEL has to stay in the prison all his life. "); } return 0; }