poj2771(二分图，最大独立集)

Guardian of Decency

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 5398 Accepted: 2269

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

2435 M classicism programming0 M baroque skiing43 M baroque chess30 F baroque soccer827 M romance programming194 F baroque programming67 M baroque ping-pong51 M classicism programming80 M classicism Paintball35 M baroque ping-pong39 F romance ping-pong110 M romance Paintball

Sample Output

37

题意：老师要找一群人漂流，可是老师怕有人谈恋爱，所以找的人之中任何两个人都不能有谈恋爱的趋势，所以给出4个没有谈恋爱趋势的标准，现在让你求老师最多带多少人去漂流。

思路：根据题目意思描述，我们要把和其他人有谈恋爱趋势最多的人找出来，所以相当于一个求最大独立集的过程，首先把有谈恋爱趋势的人连边，连完边直接用二分图跑个最大独立集就ok。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>#include<stack>using namespace std;typedef long long ll;struct data{    int h;    char sex;    char music[110],sport[110];} x[510];const int N=510;///两边的最大数量bool tu[N][N];int from[N];///记录右边的点如果配对好了它来自哪里bool use[N];///记录右边的点是否已经完成了配对int n,m;///m,n分别表示两边的各自数量,n是左边，m是右边bool dfs(int x){    for(int i=1; i<=m; i++) ///m是右边，所以这里上界是m        if(!use[i]&&tu[x][i])        {            use[i]=1;            if(from[i]==-1||dfs(from[i]))            {                from[i]=x;                return 1;            }        }    return 0;}int hungary(){    int tot=0;    memset(from,-1,sizeof(from));    for(int i=1; i<=n; i++) ///n是左边，所以这里上界是n    {        memset(use,0,sizeof(use));        if(dfs(i))            tot++;    }    return tot;}bool ok(int i,int j){    if(abs(x[i].h-x[j].h)>40)        return 1;    if(x[i].sex==x[j].sex)        return 1;    if(strcmp(x[i].music,x[j].music)!=0)        return 1;    if(strcmp(x[i].sport,x[j].sport)==0)        return 1;    return 0;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(tu,0,sizeof(tu));        scanf("%d",&n);        m=n;        for(int i=1; i<=n; i++)        {            scanf("%d %c%s%s",&x[i].h,&x[i].sex,x[i].music,x[i].sport);            for(int j=1; j<i; j++)                if(!ok(i,j))                    tu[i][j]=tu[j][i]=1;        }        printf("%d\n",n-hungary()/2);    }    return 0;}