hdu1242 Rescue bfs+优先队列

直接把Angle的位置作为起点，广度优先搜索即可，这题不是步数最少，而是time最少，就把以time作为衡量标准，加入优先队列，队首就是当前time最少的。遇到Angle的朋友就退出。只需15ms

AC代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=202;
int d[maxn][maxn];
int n,m;
char G[maxn][maxn];
struct node{
	int x,y;
	node(){
	}
	node(int x,int y):x(x),y(y){
	}
	bool operator <(const node &p)const{
		return d[x][y]>d[p.x][p.y];
	}
}s;
const int dx[]={-1,1,0,0};
const int dy[]={0,0,-1,1};
int bfs(){
	memset(d,-1,sizeof(d));
	priority_queue<node>q;
	d[s.x][s.y]=0;
	q.push(s);
	while(!q.empty()){
		node p=q.top();
		q.pop();
		int x=p.x,y=p.y;
		if(G[x][y]=='r') return d[x][y];
		for(int i=0;i<4;++i){
			int px=x+dx[i],py=y+dy[i];
			if(px<0||py<0||px>=n||py>=m||d[px][py]!=-1||G[px][py]=='#') continue;
			if(G[px][py]=='x') d[px][py]=d[x][y]+2;
			else d[px][py]=d[x][y]+1;
			q.push(node(px,py));
		}
	}
	return -1;
}
void print(){
	for(int i=0;i<n;++i){
		for(int j=0;j<m;++j)
		printf("%02d ",d[i][j]);
		printf("\n");
	}

}
int main(){
	while(scanf("%d%d",&n,&m)!=EOF){
		for(int i=0;i<n;++i)
			scanf("%s",G[i]);
		for(int i=0;i<n;++i)
		for(int j=0;j<m;++j){
			if(G[i][j]=='a') s=node(i,j);
			else if(G[i][j]!='r'&&G[i][j]!='a'&&G[i][j]!='.'&&G[i][j]!='#') G[i][j]='x';
		}
		int ans=bfs();
		//print();
		if(ans==-1) printf("Poor ANGEL has to stay in the prison all his life.\n");
		else printf("%d\n",ans);
	}
}

如有不当之处欢迎指出！