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Fire Net

Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. 

A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening. 

Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets. 

The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through. 

The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways. 



Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration. 

4
.X..
....
XX..
....
2
XX
.X
3
.X.
X.X
.X.
3
...
.XX
.XX
4
....
....
....
....
0

5
1
5
2
4

题意：

圆圈是炮台 

黑色方块是墙

炮台不能打穿墙

2个炮台摆放的位置不能在同一行或同一列 ，否则炮台会相互攻击

找出在地图上最多能放多少个炮台 

方法：深搜回溯

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,vis[6][6],maxx;
char map[6][6];
int check(int x,int y)
{
    int i;
    if(vis[x][y]!=0) return 0;
    //四个方向
    for(i=x-1; i>=0; i--)
    {
        if(vis[i][y]==1)
            break;
        else if(vis[i][y]==2)
            return 0;
    }
    for(i=x+1; i<n; i++)
    {
        if(vis[i][y]==1)
            break;
        else if(vis[i][y]==2)
            return 0;
    }
    for(i=y-1; i>=0; i--)
    {
        if(vis[x][i]==1)
            break;
        else if(vis[x][i]==2)
            return 0;
    }
    for(i=y+1; i<n; i++)
    {
        if(vis[x][i]==1)
            break;
        else if(vis[x][i]==2)
            return 0;
    }
    //printf("OK %d %d+",x,y);
    return 1;
}
int DFS(int deep)
{
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<n; j++)
        {
            if(check(i,j))
            {//printf("已经确认一个\n");
                vis[i][j]=2;//2 炮塔
                DFS(deep+1);
                vis[i][j]=0;
               // printf("已经删除一个\n");
            }
        }
    }
    if(deep>maxx)
        maxx=deep;
    return maxx;
}
int main (void)
{
    while(~scanf("%d",&n),n)
    {
        //memset(map,0,sizeof(map));
        memset(vis,0,sizeof(vis));
        getchar();
        //printf("11");
        for(int i=0; i<n; i++)
        {
            for(int ii=0; ii<n; ii++)
            {
                scanf("%c",&map[i][ii]);
                if(map[i][ii]=='X') vis[i][ii]=1;//1 强，这里输成小写x，查错好久
            }
            getchar();
        }
        maxx=0;
        printf("%d\n",DFS(0));
    }
    return 0;
}

HDU 1045 Fire Net【DFS深搜】

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原文地址：http://blog.csdn.net/qq_24653023/article/details/51897809