Simple enumeration - Bomberman 1

1. Problem Description

 We have until the bomb people bully game. Now there is a special level as follows: only one bomb, may I ask where to place this bomb can destroy most enemies?

We first model of this map, with "#" represents a wall, this wall is not blown out. Enemy with "G", said denoted by open space. "":

#############
#GG.GGG#GGG.#
###.#G#G#G#G#
#.......#..G#
#G#.###.#G#G#
#GG.GGG.#.GG#
#G#.#G#.#.###
##G...G.....#
#G#.#G###.#G#
#...G#GGG.GG#
#G#.#G#G#.#G#
#GG.GGG#G.GG#
#############

 

2. Algorithm Design 

 1. First, we use a two-dimensional array to store this map, as the bomb placed in a point which can destroy most enemies, we need one by one to try.

2. bomb blast direction is up and down four directions, so when we were enumerated for each point, the number of statistics can eliminate the need respectively along four directions. Output up to that point eventually destroy the enemy.

               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
                }

3. Source Code 

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;

const int N =111; //地图最大大小

int main()
{
    char graph[N][N];//存储地图的二维矩阵
    int n;//有多少行
    int m;//有多少列
    int p;//记录放置炸弹x坐标
    int q;//记录放置炸弹y坐标
    int sum;//记录最多消灭敌人个数
    int maxx=0;
    int x;
    int y;
    cout << "请输入地图的行n和列m" << endl;
    cin >> n >> m;
    cout << "请输入地图：" << endl;
    for(int i=0; i<=n-1; i++)
    {
        cin >> graph[i];
    }
    for(int i=0; i<=n-1; i++)
    {
        for(int j=0; j<=m-1; j++)
        {
            if(graph[i][j]=='.') //判断此时位置是否是平地
            {
                sum=0;//初始化最大消灭敌人数是0
                //将坐标i，j复制到新变量x，y中，以便上下左右统计可以消灭地人数
                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
                }
                if(sum>maxx)
                {
                    maxx=sum;
                    p=i;
                    q=j;
                }
            }
        }
    }
    cout << "将炸弹放置在" <<"("<<p<<","<<q<< ")" <<"最多可以消灭" << sum << "个敌人"<< endl;
    return 0;
}

 

4. Test results