【DBFS】——FZU2150——多起点双向广搜

                                   Problem 2150 Fire Game

Accept: 1435    Submit: 5100
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

 

题意：给出一个m*n的图，‘#’表示草坪，‘ . ’表示空地，然后可以选择在任意的两个草坪格子点火，火每 1 s会向周围四个格子扩散，

        问选择那两个点使得燃烧所有的草坪花费时间最小？

 

思路：枚举起点，双向广搜

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int INF=0x3f3f3f3f;

int dir_x[4]={1,-1,0,0};
int dir_y[4]={0,0,1,-1};

struct node
{
    int x,y;
    int t;
};

char map[15][15];
int vis[15][15];

int n,m;
int tot;//记录草方格的个数
int ans;
int run;//记录跑过的路程
bool flag=false;

queue<node> q[2];//记录分别燃烧的草地，滚动处理
node NA[120];//记录每个草方格的坐标及其步骤

bool inside(node a)
{
    if(a.x>=0&&a.x<n&&a.y>=0&&a.y<m)
        return true;
    return false;
}

void bfs(int k,int sum)//k是小孩编号，sum是以燃烧草方格个数
{
    if(sum==tot)
        flag=true;

    node u,v;
    int sin=0;//记录燃烧的草方格个数

    //由于出对元素的相邻状态入队时，是进入的另一个队列
    //因此，每清空一次处理了向外扩展一层
    while(!q[k].empty())
    {
        v=q[k].front();
        q[k].pop();

        for(int i=0;i<4;i++)
        {
            u=v;//状态转移的方法
            u.x+=dir_x[i];
            u.y+=dir_y[i];
            u.t+=1;

            if(inside(u) && map[u.x][u.y]=='#' && vis[u.x][u.y]==0)
            {
                vis[u.x][u.y]=1;
                sin++;
                run=max(run,u.t);//找出所需最大步数
                q[!k].push(u);//一次的终点作为下下次的起点
            }
        }
    }

    if(sum+sin==tot)
        flag=true;

    if(!q[!k].empty())
        bfs(!k,sum+sin);
}

int main()
{
    int t,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);

        memset(map,0,sizeof(map));
        memset(NA,0,sizeof(NA));
        memset(vis,0,sizeof(vis));

        ans=INF;
        tot=0;
        run=0;
        while(!q[0].empty()) q[0].pop();
        while(!q[1].empty()) q[1].pop();

        for(int i=0;i<n;i++)
            scanf("%s",map[i]);

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(map[i][j]=='#')
                {
                    NA[tot++]=(node){i,j,0};//struct类型的强制类型转换与赋值
                }
            }
        }

        if(tot<=2)
            ans=0;
        else
        {
            for(int i=0;i<tot;i++)
            {
                for(int j=i+1;j<tot;j++)
                {
                    //放入两个起点
                    q[0].push(NA[i]);
                    q[1].push(NA[j]);
                    memset(vis,0,sizeof(vis));

                    run=0;
                    flag=false;
                    vis[NA[i].x][NA[i].y]=vis[NA[j].x][NA[j].y]=1;//分别选中起点

                    bfs(0,2);//bfs(1,2)是一样的

                    if(flag&&ans>run)
                        ans=run;
                }
            }
        }

        if(ans==INF)
            ans=-1;
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}