把有可能在一起的连边,求最大独立点集
代码:
#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 550;
struct pupils {
int h;
string sex;
string music;
string sport;
}P[N];
int T, n;
int bmap[N][N], cy[N];
bool vis[N];
bool dfs( int u )
{
for ( int v = 1; v <= n; ++v ) if ( bmap[u][v] && !vis[v] ) {
vis[v] = 1;
if ( cy[v] == -1 || dfs(cy[v])) {
cy[v] = u;
return 1;
}
}
return 0;
}
int match()
{
int res = 0;
memset( cy, -1, sizeof(cy) );
for ( int i = 1; i <= n; ++i ) {
memset( vis, 0, sizeof(vis));
if ( dfs(i)) res++;
}
return res;
}
int main()
{
scanf("%d", &T);
while ( T-- ) {
scanf("%d", &n);
for ( int i = 1; i <= n; ++i ) {
cin >> P[i].h >> P[i].sex >> P[i].music >> P[i].sport;
//cout << P[i].h << " " << P[i].sex << " " << P[i].music << " " << P[i].sport << endl;
}
memset( bmap, 0, sizeof(bmap));
for ( int i = 1; i <= n; ++i )
for ( int j = 1+i; j <= n; ++j ) if ( abs(P[i].h-P[j].h) <= 40 && P[i].sex != P[j].sex && P[i].music == P[j].music && P[i].sport != P[j].sport )
bmap[i][j] = bmap[j][i] = 1;
printf("%d\n", n-match()/2);
}
}