E - Fire Game——双起点BFS

小咪咪 2022-06-16 14:24 131阅读 0赞

Think:  
1双起点BFS  
2队列思想（结构体队列）实现BFS

E - Fire Game 
    
    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

以下为Accepted代码

#include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    struct node
    {
        int x;
        int y;
        int z;
    }link[144], e[144];
    //link数组实现队列思想
    //e数组记录符合条件的点的集合
    
    int n, m, op, tp, t;
    int v[14][14];//v数组记录当前结点是否入队过
    char st[14][14];//st数组记录原始输入信息
    int Move[4][2] = {
       {
       0, 1}, {
       1, 0}, {
       0, -1}, {-1, 0}};//Move数组实现当前结点的不同方向的移动
    
    int Ans();//判断能否从当前位置遍历所有符合条件的结点，若能且返回遍历所有符合条件的结点所需时间
    bool Judge(int dx, int dy);//判断当前结点是否在原始输入信息中
    
    int main()
    {
        int T, i, j, k;
        scanf("%d", &T);
        for(k = 1; k <= T; k++)
        {
            t = 0;
            printf("Case %d: ", k);
            scanf("%d %d", &n, &m);
            for(i = 0; i < n; i++)
                scanf("%s", st[i]);
    
            for(i = 0; i < n; i++){
                for(j = 0; j < m; j++){
                    if(st[i][j] == '#'){
                        e[t].x = i, e[t].y = j, e[t].z = 0;
                        t++;
                    }
                }
            }
    
            int s = 144;
            for(i = 0; i < t; i++){
                for(j = i; j < t; j++){
                    op = tp = 0;//队列初始化
                    memset(v, 0, sizeof(v));//记录当前结点是否入队过的数组的初始化
                    v[e[i].x][e[i].y] = 1;//标记初始结点入队
                    v[e[j].x][e[j].y] = 1;//标记初始结点入队
                    link[tp++] = e[i];//初始结点入队
                    link[tp++] = e[j];//初始结点入队
                    s = min(s, Ans());//判断当前情况的优解情况
                }
            }
            if(s == 144)//不能遍历所有的符合条件的结点
                printf("-1\n");
            else//能够遍历所有的符合条件的结点
                printf("%d\n", s);
        }
        return 0;
    }
    bool Judge(int dx, int dy)//判断当前结点是否在原始输入数据中
    {
        if(dx < 0 || dx >= n || dy < 0 || dy >= m)
            return false;
        else
            return true;
    }
    int Ans()//判断能否从当前位置遍历所有符合条件的结点，若能且返回遍历所有符合条件的结点所需时间
    {
        int i, z;
        z = link[op].z;//记录最初时间
        while(op < tp)
        {
            for(i = 0; i < 4; i++){
       //从当前结点开始进行移动
                int x = link[op].x+Move[i][0];
                int y = link[op].y+Move[i][1];
                if(Judge(x, y) && st[x][y] == '#' && v[x][y] == 0){
       //判断是否符合条件
                    v[x][y] = 1;//标记当前结点入队
                    link[tp].x = x;//当前结点入队
                    link[tp].y = y;//当前结点入队
                    link[tp].z = link[op].z + 1;//当前结点入队
                    z = max(z, link[tp].z);//所有结点中时间遍历的最大值即遍历所有符合条件的结点的所需时间
                    tp++;
                }
            }
            op++;
        }
        if(op < t)//不能遍历所有符合条件的结点
            return 144;
        else//能遍历所有符合条件的结点，且返回所需时间
            return z;
    }