HDU - 1072 Nightmare

题意：求走出迷宫的最短时间，2代表起点，3代表出口，每次经过4都可以将爆炸时间归0，要在爆炸时间内出去

思路：这里的BFS,vis[][]不再代表是否走过而是距离爆炸的时间，还有就是可以重复的经过一个点，所以我们再次走到这个点的时候，如果能有更长的爆炸时间就入队

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int MAXN = 20;

struct node{
	int x,y,t,step;
}tmp;
queue<node> q;
int dx[] = {-1,1,0,0};
int dy[] = {0,0,-1,1};
int map[MAXN][MAXN],vis[MAXN][MAXN];
int n,m,x1,y1,ans;

void bfs(){
	while (!q.empty()){
		tmp = q.front();
		q.pop();
		int x = tmp.x,y = tmp.y,t = tmp.t - 1,step = tmp.step + 1;
		if (tmp.t == 0)
			continue;
		else {
			if (map[x][y] == 3){
				ans = tmp.step;
				return;
			}
			if (map[x][y] == 4)
				t = 5;
		}
		for (int i = 0; i < 4; i++){
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (nx >= 0 && nx < n && ny >= 0 && ny < m && vis[nx][ny] < t && map[nx][ny] != 0){
				vis[nx][ny] = t;
				tmp.x = nx,tmp.y = ny;
				tmp.t = t,tmp.step = step;
				q.push(tmp);
			}
		}
	}
}

int main(){
	int t;
	scanf("%d",&t);
	while (t--){
		scanf("%d%d",&n,&m);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++){
				scanf("%d",&map[i][j]);
				vis[i][j] = 0;
				if (map[i][j] == 2)
					x1 = i,y1 = j;
			}
		while (!q.empty())
			q.pop();
		ans = -1,vis[x1][y1] = 6;
		tmp. x = x1,tmp.y = y1,tmp.t = 6,tmp.step = 0;
		q.push(tmp);
		bfs();
		printf("%d\n",ans);
	}
	return 0;
}

HDU - 1072 Nightmare,古老的榕树,5-wow.com