UVa 11624 Fire!（起火迷宫）

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题目链接：UVa 11624

题意：

迷宫中有着火点，每次向四周扩散（墙不会被点燃），问能否逃出？若逃出需几步？

分析：

一开始是想用bfs先记录下着火的顺序，也就是可走的方格在哪一步被点着，然后在用bfs找最短路径时多个判断即可。可是这样就TLE了，然后就废了好大劲，将两个bfs合并了，难点就是return条件判断和“剪枝”。从早上做到现在，终于AC了！

CODE:

#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <queue>#include <algorithm>#include <set>using namespace std;const int maxn = 1005;int dir[4][2] = { {-1,0},{0,-1},{1,0},{0,1} };int R, C, T, ans, jx, jy, fx, fy;int wall[maxn][maxn];char s[maxn], a[maxn][maxn];int vis[maxn][maxn];int escapex, escapey, firex, firey;int fire[maxn][maxn];int valid(int x, int y){if (x < 0 || y < 0 || x == R || y == C || wall[x][y]) return 0;return 1;}struct Node {int step, x, y;int isfire;}cur, nextnode;int bfs(){queue<Node> q;memset(vis, 0, sizeof(vis));memset(fire, 0, sizeof(fire));for (int i = 0;i < R;i++){for (int j = 0;j < C;j++){if (a[i][j] == 'F'){cur.x = i;cur.y = j;cur.step = 0;cur.isfire = 1;fire[i][j] = 1;q.push(cur);//起火点先进队列}}}cur.x = jx;cur.y = jy;cur.step = 0;cur.isfire = 0;vis[jx][jy] = 1;q.push(cur);//人后进队列while (!q.empty()){cur = q.front();q.pop();if (cur.isfire){for (int i = 0;i < 4;i++){firex = cur.x + dir[i][0];firey = cur.y + dir[i][1];if (!valid(firex, firey) || fire[firex][firey]) continue;//fire[i][j]:坐标（i，j）已着火nextnode.step = cur.step + 1;nextnode.x = firex;nextnode.y = firey;nextnode.isfire = 1;fire[firex][firey] = 1;q.push(nextnode);}}else{for (int i = 0;i < 4;i++){escapex = cur.x + dir[i][0];escapey = cur.y + dir[i][1];nextnode.step = cur.step + 1;nextnode.x = escapex;nextnode.y = escapey;nextnode.isfire = 0;if (nextnode.x == -1 || nextnode.x == R || nextnode.y == -1 || nextnode.y == C)return nextnode.step;if (!valid(escapex,escapey)|| vis[escapex][escapey] || fire[escapex][escapey]) continue;vis[escapex][escapey] = 1;q.push(nextnode);}}}return -1;}int main(){#ifdef LOCAL freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endifcin >> T;while (T--){cin >> R >> C;memset(wall, 0, sizeof(wall));for (int i = 0;i < R;i++){scanf("%s", s);for (int j = 0;j < C;j++){a[i][j] = s[j];if (s[j] == '#') wall[i][j] = 1;else if (s[j] == 'J'){jx = i;jy = j;}}}ans = bfs();if (ans == -1) printf("IMPOSSIBLE\n");else printf("%d\n", ans );}return 0;}

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