[kuangbin带你飞]专题一 简单搜索 -I - Fire Game

 Problem 2150 Fire Game

Accept: 1352    Submit: 4789
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows, M columns). At the beginning, each grid of this board is consisting of grass or just empty and then they start to fire all the grass. Firstly they choose two grids which are consisting of grass and set fire. As we all know, the fire can spread among the grass. If the grid (x, y) is firing at time t, the grid which is adjacent to this grid will fire at time t+1 which refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends when no new grid get fire. If then all the grid which are consisting of grass is get fired, Fat brother and Maze will stand in the middle of the grid and playing a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.)

You can assume that the grass in the board would never burn out and the empty grid would never get fire.

Note that the two grids they choose can be the same.

 Input

The first line of the date is an integer T, which is the number of the text cases.

Then T cases follow, each case contains two integers N and M indicate the size of the board. Then goes N line, each line with M character shows the board. “#” Indicates the grass. You can assume that there is at least one grid which is consisting of grass in the board.

1 <= T <=100, 1 <= n <=10, 1 <= m <=10

 Output

For each case, output the case number first, if they can play the MORE special (hentai) game (fire all the grass), output the minimal time they need to wait after they set fire, otherwise just output -1. See the sample input and output for more details.

 Sample Input

4 

3 3

.#.

###

.#.

3 3

.#.

#.#

.#.

3 3

...

#.#

...

3 3

###

..#

#.#

 Sample Output

Case 1: 1

Case 2: -1

Case 3: 0

Case 4: 2

分析：首先，判断题目的草丛数，小于等于2的直接剪枝。

其次，枚举两堆草丛燃烧时间并记录最小值。

注意：燃烧事件应该以两个草丛中最长的为准。多于两个的直接－1。。

code:[718ms]

#include <cstdio>
#include <queue>
#include <iostream>
#include <cstring>

using namespace std;
typedef long long ll;
typedef struct {
    int x;
    int y;
    int wal;
}data;
data yihao,erhao;
data map_temp[105];

int n, m,cas, vis[15][15];
int dir[][2] = { {0,1} , {1,0} , {0,-1} , {-1,0}  };
char grid[15][15];

void init()
{
    memset(grid, 0, sizeof grid);
    for (int i = 1; i <= n; i++)
        scanf("%s", grid[i] + 1);
}

int bfs(int x1,int y1,int x2,int y2)
{
    int ans = 0;
    memset(vis, 0, sizeof vis);
    vis[x1][y1] = 1;
    vis[x2][y2] = 1;
    
    yihao.x = x1, yihao.y = y1;
    erhao.x = x2, erhao.y = y2;
    yihao.wal = erhao.wal = 0;
    queue <data> q;
    q.push(yihao);
    q.push(erhao);
    
    while (q.size())
    {
        data now = q.front(); q.pop();
        data nex;
        
        if (ans < now.wal)
        {
            ans = now.wal;
        }
        for (int i = 0; i < 4; i++)
        {
            int a = now.x + dir[i][0], b = now.y + dir[i][1];
            if (vis[a][b] == 0&&1<=a&&a<=n&&1<=b&&b<=m)
                if(grid[a][b]=='#')
                {
                    nex.x = a;
                    nex.y = b;
                    nex.wal = now.wal + 1;
                    q.push(nex);
                    vis[a][b] = 1;
                }
        }
    }
    return ans;
}

int cheack()
{

    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(grid[i][j]=='#')
            {
                if(vis[i][j]==0)
                {
                    return 0;
                }
            }
        }
    return 1;
}

void solve()
{
    int ans = 1000000;
    int cnt = 0;
    for(int i=1;i<=n;i++)
        for (int j = 1; j <= m; j++)
        {
            if (grid[i][j] == '#')
            {
                map_temp[cnt].x=i,map_temp[cnt++].y=j;
            }
        }
    
    if (cnt <= 2)
    {
        printf("Case %d: 0\n", cas++);
        return;
    }
    
    int temp;
    for (int i = 0; i < cnt; i++)
    {
        for (int j = 0; j < cnt; j++)
        {
            if(i==j) continue;
            temp=bfs(map_temp[i].x,map_temp[i].y,map_temp[j].x,map_temp[j].y);
            if(cheack())
                if(ans>temp)
                    ans=temp;
        }
    }
    
    if(ans==1000000)
        printf("Case %d: -1\n",cas++);
    else
        printf("Case %d: %d\n", cas++, ans);
}

int main(void)
{
    cas = 1;
    int t;
    cin >> t;
    while (t--)
    {
        scanf("%d %d",&n,&m);
        init();
        solve();
    }
}

quesiton:双向bfs优化的实现..