Nightmare（HDU 1072）

题目大意：一般的搜索问题，用BFS能够0ms解决。有剩余时限让你去走迷宫，有一个特殊的东西可以让你的剩余时间重置。

具体的要求是：

1. We can assume the labyrinth is a 2 array.(二维迷宫)
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.（不能走墙）
3. If Ignatius get to the exit when the exploding time turns to 0, he can't get out of the labyrinth.（刚好在剩余0S走出也不行）
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can't use the equipment to reset the bomb.（刚好在0s重置时间也不行）
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.（你可以使用重置若干次，不过在我的代码当中，是没考虑，不知道是不是测试数据太弱了，就那么AC了。）
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.（一走到该道具，剩余时间立刻重置，不需要考虑时间）

测试数据：

33 32 1 11 1 01 1 34 82 1 1 0 1 1 1 01 0 4 1 1 0 4 11 0 0 0 0 0 0 11 1 1 4 1 1 1 35 81 2 1 1 1 1 1 4 1 0 0 0 1 0 0 1 1 4 1 0 1 1 0 1 1 0 0 0 0 3 0 1 1 1 4 1 1 1 1 1 

4-113

以下是我的AC代码：一般的BFS模板根据条件改编

#include<iostream>#include <queue>#include <cstring>using namespace std;short map[10][10], vis[10][10];int Div[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };int n, m;int sx, sy, ex, ey;int times = -1;struct pos{int x;int y;int times;int Time;friend bool operator <(pos a, pos b)//使用优先队列，使得剩余时间最多的先出队{return a.Time<b.Time;}};void bfs(){memset(vis, 0, sizeof (vis));priority_queue<pos>q;pos p, mid;p.x = sx;p.y = sy;p.times = 0;p.Time = 6;q.push(p);vis[sx][sy] = 1;while (!q.empty()){p = q.top();q.pop();if (p.x == ex&&p.y == ey&&p.Time > 0){times = p.times;return;}if (p.Time == 1)//这个是针对0秒不可行，即当剩余时间为1S是还没走出，那永远不可以。continue;for (int i= 0; i < 4; i++)//这边加入一个判断，是为了判断该点附近有没有时间重置的地点。{mid = p;//具体理解看第三个测试数据吧，还好测试数据给的良心，不然完全不会想到。mid.x = p.x + Div[i][0];//道行还是太浅了。mid.y= p.y + Div[i][1];if (mid.x >= 1 && mid.x <= n&&mid.y <= m&&mid.y >= 1 && map[mid.x][mid.y] == 4&&p.Time<5){mid.Time = 6;mid.times++;//本来想解释一下，但发现不会解释，看不懂的还是看测试数据三理解吧q.push(mid);p.Time = 5;p.times += 2;break;}}for (int i = 0; i < 4; i++){mid = p;mid.x = p.x + Div[i][0];mid.y = p.y + Div[i][1];mid.times++;mid.Time--;if (mid.x>n || mid.x<1 || mid.y>m || mid.y < 1||!map[mid.x][mid.y]||vis[mid.x][mid.y])continue;if (map[mid.x][mid.y] == 4){vis[mid.x][mid.y] = 1;mid.Time = 6;q.push(mid);continue;}else{vis[mid.x][mid.y] = 1;q.push(mid);}}}}int main(){int t;cin >> t;while (t--){cin >> n >> m;times = -1;for (int i = 1; i <= n;i++)for (int j = 1; j <= m; j++){cin >> map[i][j];if (map[i][j] == 2){sx = i; sy = j;}if (map[i][j] == 3){ex = i; ey = j;}}bfs();cout << times << endl;}}