POJ 2771 Guardian of Decency(二分图最大独立集)

题意：老师带学生出去旅游，但是担心学生会发生恋爱关系，所以带出去的学生至少要满足以下要求之中的一个：

           1.身高相差40cm以上

           2.同性

           3.喜欢的音乐风格不同

           4.喜欢的运动相同

           问最多可以带出去多少学生？

思路：显然的一个二分图匹配，左边为男同学，右边为女同学，可以如果建图的时候正着建会很困难，所以我们反着建图，如果他们两不满足以上4个条件的任何一条,那么在这两个学生之间连一条边.这样我们就建立了一个二分图.(如果存在边肯定是左边与右边之间的边)，那么该图的最大独立集中的任意两点就一定满足上面4个条件的至少一个

#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<iostream>
#include<cmath>
using namespace std;
const int maxn=1000;

struct Max_Match
{
    int n,m;
    vector<int> g[maxn];
    bool vis[maxn];
    int left[maxn];
    void init(int n)
    {
        this->n=n;
      //  this->m=m;
        for(int i=0;i<=n;i++) g[i].clear();
        memset(left,-1,sizeof(left));
    }
    bool match(int u)
    {
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            if(!vis[v])
            {
                vis[v]=true;
                if(left[v]==-1 || match(left[v]))
                {
                    left[v]=u;
                    return true;
                }
            }
        }
        return false;
    }
    
    int solve()
    {
        int ans=0;
        for(int i=0;i<n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(match(i)) ans++;
        }
        return ans;
    }
}MM;
int T;
struct Node
{
	double height;
	string music;
	string sport;
	string sex;
}man[maxn],girl[maxn];

int main()
{
	int n,m;
	scanf("%d",&T);
	while (T--)
	{
		int boy = 0;
		int gir = 0;
		scanf("%d",&n);
		for (int i = 1;i<=n;i++)
		{
            double temp;
			string sex;
			string mu,sp;
			cin >> temp >> sex >> mu >> sp;
		    if (sex == "M")
			{
				man[boy].sex=sex;
                man[boy].height = temp;
                man[boy].music = mu;
				man[boy++].sport = sp;
			}	
			else
			{
				girl[gir].sex=sex;
				girl[gir].height = temp;
				girl[gir].music = mu;
				girl[gir++].sport = sp;
			}
		}
        MM.init(boy);
		for (int i = 0;i<boy;i++)
			for (int j = 0;j<gir;j++)
			{
				if (abs(man[i].height-girl[j].height)<=40 && (man[i].music == girl[j].music)&&(man[i].sport!=girl[j].sport))
					MM.g[i].push_back(j);
			}
		printf("%d\n",n-MM.solve());
	}
}

Description

Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made some rules that he thinks indicates a low probability two persons will become a couple: 

• Their height differs by more than 40 cm. 
  
• They are of the same sex. 
  
• Their preferred music style is different. 
  
• Their favourite sport is the same (they are likely to be fans of different teams and that would result in fighting).

So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information. 

Input

The first line of the input consists of an integer T ≤ 100 giving the number of test cases. The first line of each test case consists of an integer N ≤ 500 giving the number of pupils. Next there will be one line for each pupil consisting of four space-separated data items: 

• an integer h giving the height in cm; 
  
• a character 'F' for female or 'M' for male; 
  
• a string describing the preferred music style; 
  
• a string with the name of the favourite sport.

No string in the input will contain more than 100 characters, nor will any string contain any whitespace. 

Output

For each test case in the input there should be one line with an integer giving the maximum number of eligible pupils.

Sample Input

2
4
35 M classicism programming
0 M baroque skiing
43 M baroque chess
30 F baroque soccer
8
27 M romance programming
194 F baroque programming
67 M baroque ping-pong
51 M classicism programming
80 M classicism Paintball
35 M baroque ping-pong
39 F romance ping-pong
110 M romance Paintball

Sample Output

3
7