Guardian of Decency
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 4071 | Accepted: 1702 |
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).
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.
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
Source
二分匹配。
主要是建图方法。
题目要求是选出两两之间满足四个条件之一的人。
我们建图应该反过来建。
给不满足四个条件的人一条边。就转换成求最大独立集的问题。
二分图最大独立集=顶点数-二分图最大匹配
独立集：图中任意两个顶点都不相连的顶点集合。
#include#include #include #include #include using namespace std;/* **************************************************************************//二分图匹配（匈牙利算法的DFS实现）//初始化：g[][]两边顶点的划分情况//建立g[i][j]表示i->j的有向边就可以了，是左边向右边的匹配//g没有边相连则初始化为0//uN是匹配左边的顶点数，vN是匹配右边的顶点数//调用：res=hungary();输出最大匹配数//优点：适用于稠密图，DFS找增广路，实现简洁易于理解//时间复杂度:O(VE)//***************************************************************************///顶点编号从0开始的const int MAXN=510;int uN,vN;//u,v数目int g[MAXN][MAXN];int linker[MAXN];bool used[MAXN];bool dfs(int u)//从左边开始找增广路径{ int v; for(v=0;v