BFS(两点搜索) FZOJ 2150 Fire Game

题目传送门

题意：'#'表示草地，两个人在草地上点火，相邻的草地会烧起来，每烧一格等1秒，问最少要等几秒草地才烧完

分析：这题和UVA 11624 Fire!有点像，那题给定了两个点，这题两点不确定，取最小值。可以仿照11624的做法，两次BFS，第二次更新最小值，这样我跑了900多ms。后来发现不需要这样，枚举两个点，将它们一起入队，vis过的点一定已经是最优的，广搜的特点。

收获：1. 进一步理解BFS的过程 2. 差错能力有待提高 3. 写过的题一定要总结！(刚刷过专题！！！)

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-23 12:40:49
* File Name     :J.cpp
 ************************************************/
 
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
 
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 12;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct Node	{
	int x, y, step;
};
Node g[N*N];
char maze[N][N];
int d[N][N];
bool vis[N][N];
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int n, m, cnt, res;
 
bool judge(int x, int y)	{
	if (x < 1 || x > n || y < 1 || y > m || vis[x][y] || maze[x][y] != '#')	return false;
	return true;
}
 
void BFS(int sx, int sy, int ex, int ey)	{
	queue<Node> Q;
    Q.push ((Node) {sx, sy, 0});
    Q.push ((Node) {ex, ey, 0});
	vis[sx][sy] = vis[ex][ey] = true;
    while (!Q.empty ())	{
		Node r = Q.front ();	Q.pop ();
		int x = r.x, y = r.y, step = r.step;
        res = max (res, step);
		for (int i=0; i<4; ++i)	{
			int tx = x + dx[i], ty = y + dy[i];
			if (!judge (tx, ty))	continue;
            vis[tx][ty] = true;
            Q.push ((Node) {tx, ty, step + 1});
		}
	}
}
 
bool check(void)	{
	for (int i=1; i<=cnt; ++i)	{
		int x = g[i].x, y = g[i].y;
		if (!vis[x][y])	return false;
	}
    return true;
}
 
int work(void)	{
	int ans = INF;
    if (cnt <= 2)   return 0;
	for (int i=1; i<=cnt; ++i)	{
		for (int j=i+1; j<=cnt; ++j)	{
			memset (vis, false, sizeof (vis));  res = 0;
			BFS (g[i].x, g[i].y, g[j].x, g[j].y);
			if (check ())   ans = min (ans, res);
		}
	}
 
	if (ans == INF)	return -1;
	else	return ans;
}
 
int main(void)    {
	int T, cas = 0;	scanf ("%d", &T);
	while (T--)	{
		cnt = 0;
		scanf ("%d%d", &n, &m);
		for (int i=1; i<=n; ++i)	{
			scanf ("%s", maze[i] + 1);
			for (int j=1; j<=m; ++j)	{
				if (maze[i][j] == '#')	{
					g[++cnt].x = i;	g[cnt].y  = j;
				}
			}
		}
 
		printf ("Case %d: %d\n", ++cas, work ());
	}
 
    return 0;
}