DFS-- Bomberman 3

1. Analysis

Continue on the contents of both the blog, on Bomberman, we can also be solved by DFS, DFS explained in detail in the examples:

https://blog.csdn.net/Jayphone17/article/details/102702960

Code is also similar, reference may be

2. Algorithm Design 

 Right away from the start point. Whatever new statistics point to point can destroy the enemy, and to continue to try to go down from that point, until no way to go to return, and then try the other direction, until all the points you can go visit again, deep search ended.

3. Source Code 

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=111;
char graph[N][N];
int marked[N][N];
int maxx;
int n;
int m;
int mx;
int my;

int getnum(int i,int j)//返回最大消灭敌人数的函数
{
    int sum=0;
    int x;
    int y;
    x=i;
    y=j;
    while(graph[x][y]!='#')//只要没有到边界
    {
        if(graph[x][y]=='G')//是敌人
        {
            sum++;//增加消灭数
        }
        x--;//继续向上寻找
    }
    x=i;
    y=j;
    while(graph[x][y]!='#')//只要没有到边界
    {
        if(graph[x][y]=='G')//是敌人
        {
            sum++;//增加消灭数
        }
        x++;//继续向上寻找
    }
    x=i;
    y=j;
    while(graph[x][y]!='#')//只要没有到边界
    {
        if(graph[x][y]=='G')//是敌人
        {
            sum++;//增加消灭数
        }
        y--;//继续向上寻找
    }
    x=i;
    y=j;
    while(graph[x][y]!='#')//只要没有到边界
    {
        if(graph[x][y]=='G')//是敌人
        {
            sum++;//增加消灭数
        }
        y++;//继续向上寻找
    }
    return sum;
}

void DFS(int x,int y)
{
    int next[4][2]= {{0,1},{1,0},{0,-1},{-1,0}}; //定义四个方向的数组
    int tx;
    int ty;
    int sum;
    sum=getnum(x,y);
    if(sum>maxx)
    {
        maxx=sum;
        mx=x;
        my=y;
    }
    for(int k=0; k<=3; k++)
    {
        tx=x+next[k][0];
        ty=y+next[k][1];
        if(tx<1 || tx>n-1 || ty<1 || ty>m-1)
        {
            continue;
        }
        if(graph[tx][ty]=='.' && marked[tx][ty]==0)
        {
            marked[tx][ty]=1;
            DFS(tx,ty);
        }
    }
    return ;
}

int main()
{
    int startx;
    int starty;
    cout << "请输入地图的行数n和列数m以及起点的坐标"<< endl;
    cin >> n>> m>> startx>>starty;
    cout << "请输入地图"<< endl;
    for(int i=0; i<=n-1; i++)
    {
        cin >> graph[i];
    }
    marked[startx][starty]=1;
    maxx=getnum(startx,starty);
    mx=startx;
    my=starty;
    DFS(startx,starty);
    cout << "将炸弹放在" << "(" << mx << "," << my << ")"<< "处了以消灭最多敌人：" << maxx << endl;
    return 0;
}

 

4. Test results