Battle ships(二分图，建图，好题)

Battle ships

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1007    Accepted Submission(s): 353

Problem Description

Dear contestant, now you are an excellent navy commander, who is responsible of a tough mission currently.
Your fleet unfortunately encountered an enemy fleet near the South Pole where the geographical conditions are negative for both sides. The floating ice and iceberg blocks battleships move which leads to this unexpected engagement highly dangerous, unpredictable and incontrollable. 
But, fortunately, as an experienced navy commander, you are able to take opportunity to embattle the ships to maximize the utility of cannons on the battleships before the engagement. 
The target is, arrange as many battleships as you can in the map. However, there are three rules so that you cannot do that arbitrary:
A battleship cannot lay on floating ice A battleship cannot be placed on an iceberg
Two battleships cannot be arranged in the same row or column, unless one or more icebergs are in the middle of them.

 

Input

There is only one integer T (0<T<12) at the beginning line, which means following T test cases.
For each test case, two integers m and n (1 <= m, n <= 50) are at the first line, represents the number of rows and columns of the battlefield map respectively. Following m lines contains n characters iteratively, each character belongs to one of ‘#’, ‘*’, ‘o’, that symbolize iceberg, ordinary sea and floating ice.

 

Output

For each case, output just one line, contains a single integer which represents the maximal possible number of battleships can be arranged.

 

Sample Input

2
4 4
*ooo
o###
**#*
ooo*
4 4
#***
*#**
**#*
ooo#

 

Sample Output

3
5

 

题解：题意就是让找在这个矩阵中可以放的船最大数目；其中船之间可以有冰山，船间没冰山时船不能在同行同列；这样需要建图，先扫描行，如果遇到冰山t++;下一个数字可以继续匹配，下一行了t++,同样，再扫描列；得到二分匹配的x分部和y分部；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define T_T while(T--)
#define F(i,x) for(i=0;i<x;i++)
#define PR(x) printf("%d",x)
#define PL(x) printf("%lld",x)
#define p_ printf(" ")
const int MAXN=1010;
char s[MAXN][MAXN];
int mp[MAXN][MAXN],vis[MAXN],link[MAXN],x[110][110],y[110][110];
int m,n,t1,t2;
int getmpxy(int rl,int a[][110]){//扫描行和列得到二分匹配的x部和y部
	int t=0,i,j;
	if(rl)F(i,m){
		F(j,n){
			if(s[i][j]=='*')a[i][j]=t;
			if(s[i][j]=='#')t++;
		}
		t++;
	}
	else F(j,n){
		F(i,m){
			if(s[i][j]=='*')a[i][j]=t;
			if(s[i][j]=='#')t++;
		}
		t++;
	}
	return t;
}
void getmp(){
	mem(x,0);mem(y,0);
	t1=getmpxy(1,x);
	t2=getmpxy(0,y);
	int i,j;
	F(i,m){
		F(j,n){
			if(s[i][j]=='*'){
			//	printf("%d %d\n",x[i][j],y[i][j]);
				mp[x[i][j]][y[i][j]]=1;
			}
		}
	}
}
bool dfs(int x){
	int i,j;
	F(i,t2){
		if(!vis[i]&&mp[x][i]){
			vis[i]=1;
			if(link[i]==-1||dfs(link[i])){//||
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}
int km(){
	mem(link,-1);
	int i,j;
	int ans=0;
	F(i,t1){
		mem(vis,0);
		if(dfs(i))ans++;
	}
	return ans;
}
int main(){
	int T;
	SI(T);
	T_T{
		SI(m);SI(n);
		int i,j;
		mem(mp,0);
		F(i,m)
		scanf("%s",s[i]);
		getmp();
		PR(km());
		puts("");
	}
	return 0;
}