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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

Zhejiang University Local Contest 2001 

DFS深度优先还得多练习啊，下边还有单靠大神的代码用二分图写的代码，第一次用，，只怪自己太菜

//DFS 
#include<stdio.h>
#include<string.h>
char mp[8][8];
int n,ans;
int check(int x,int y){
	int i;
	//判断行是否冲突 
	for(i=y-1;i>0;i--){
		if(mp[x][i]=='#') return 0;
		if(mp[x][i]=='X') break;
	}
	for(i=y+1;i<=n;i++){
		if(mp[x][i]=='#') return 0;
		if(mp[x][i]=='X') break;
	}
	//判断列是否冲突 
	for(i=x-1;i>0;i--){
		if(mp[i][y]=='#') return 0;
		if(mp[i][y]=='X') break;
	}
	for(i=x+1;i<=n;i++){
		if(mp[i][y]=='#') return 0;
		if(mp[i][y]=='X') break;
	}
	return 1;
}

void dfs(int cnt){
	if(cnt>ans) ans=cnt;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			if(mp[i][j]=='.'&&check(i,j)){
				mp[i][j]='#';
				dfs(cnt+1);
				mp[i][j]='.';
			}
		}
} 
int main(){
	int i,j;
	while(scanf("%d",&n),n){
		for(i=1;i<=n;i++){
			getchar();
			for(j=1;j<=n;j++){
				scanf("%c",&mp[i][j]);
			}
		}
		ans=0;
		dfs(0);
		printf("%d\n",ans);
	}
	return 0;
}
//下边是用二分图匹配写的代码
#include<stdio.h>
#include<string.h>
char mp[8][8];
int r[8][8],c[8][8],path[8][8];
int n,rmax,cmax,ruse[8],cuse[8],vis[8];
int Find(int x){
	for(int i=1;i<cmax;i++){//在列集合中找到有路径的点 
		if(!vis[i] && path[x][i]){
			vis[i]=1;
			if(!cuse[i] || Find(cuse[i])){	
				cuse[i]=x;//分别赋值 
				ruse[x]=i;
				return 1;
			}
		}
	}
	return 0;
}
int main(){
	int i,j;
	while(scanf("%d",&n),n){
		for(i=1;i<=n;i++){
			getchar();
			for(j=1;j<=n;j++){
				scanf("%c",&mp[i][j]);
			}
		}
		memset(r,-1,sizeof(r));
		memset(c,-1,sizeof(c));
		//缩点 
		rmax=cmax=1;
		for(i=1;i<=n;i++){
			for(j=1;j<=n;j++){
				if(mp[i][j]=='.'&&r[i][j]==-1){//横向缩点 
					for(int k=j;k<=n&&mp[i][k]=='.';k++)
						r[i][k]=rmax;//相同区域标记同一个数字 
					rmax++;
				}
				if(mp[j][i]=='.'&&c[j][i]==-1){//纵向缩点 
					for(int k=j;k<=n&&mp[k][i]=='.';k++)
						c[k][i]=cmax;//相同区域标记同一个数字 
					cmax++;
				}
			}
		}
		//连边建立路径图 
		memset(path,0,sizeof(path));
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++){
				if(mp[i][j]=='.'){
					path[ r[i][j] ][ c[i][j] ]=1;
				}
			}
		//二分图匹配
		int ans=0;
		memset(ruse,0,sizeof(ruse));
		memset(cuse,0,sizeof(cuse));
		for(i=1;i<rmax;i++) {// 行集合中搜索 
			memset(vis,0,sizeof(vis));
			ans+=Find(i);
		}
		
		printf("%d\n",ans);
	}
	return 0;
}

hdu 1045 && zoj 1002 Fire Net（DFS && 二分图匹配）

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原文地址：http://blog.csdn.net/ling_du/article/details/47343027