poj 2771 二分图最大独立集

最新推荐文章于 2025-04-21 10:23:33 发布

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最新推荐文章于 2025-04-21 10:23:33 发布

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本文介绍了一道关于二分图最大独立集的问题，并提供了一个具体的AC代码示例。该问题关注如何根据学生的特定属性（如身高、性别、音乐喜好和喜爱运动），确保选出的学生中任意两人至少有一项不符合情侣条件，以最大化出游学生数量。

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是一道二分图最大独立集的裸题，，，就是看着麻烦了点，，其实很简单。题目：

Guardian of Decency

Time Limit:3000MS		Memory Limit:65536K
Total Submissions:3584		Accepted:1493

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

ac代码：

#include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <string.h> #include <vector> #include <string> using namespace std; vector<int> ss[505]; int flag[505],visted[505]; struct people{ int height; char male; string music; string pe; }aa[505]; bool dfs(int x){ for(int i=0;i<ss[x].size();++i){ if(!visted[ss[x][i]]){ visted[ss[x][i]]=1; if(flag[ss[x][i]]==-1||dfs(flag[ss[x][i]])){ flag[ss[x][i]]=x; return true; } } } return false; } int main(){ int t; scanf("%d",&t); while(t--){ int n; memset(flag,-1,sizeof(flag)); memset(ss,0,sizeof(ss)); scanf("%d",&n); for(int i=0;i<n;++i){ cin>>aa[i].height>>aa[i].male>>aa[i].music>>aa[i].pe; } for(int i=0;i<n;++i){ for(int j=0;j<n;++j){ int x=abs(aa[j].height-aa[i].height); if(x<=40&&aa[j].male!=aa[i].male&&aa[j].music==aa[i].music&&aa[j].pe!=aa[i].pe){ ss[i].push_back(j); } } } int sum=0; for(int i=0;i<n;++i){ memset(visted,0,sizeof(visted)); if(dfs(i)) sum++; } printf("%d\n",n-sum/2); } return 0; }