【hdu 1068】Girls and Boys

【Link】:http://acm.hdu.edu.cn/showproblem.php?pid=1068

【Description】



有n个人，一些人认识另外一些人，选取一个集合，使得集合里的每个人都互相不认识，求该集合中人的最大个数。

【Solution】



最大独立子集问题;

等于节点个数-最小点覆盖.

写之前做一下染色,从0开始写最大匹配.



【NumberOf WA】



0



【Reviw】

【Code】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x+1)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1000;

int n,color[N+10],Try[N+10],pre[N+10];
vector <int> v[2],g[N+10];

void dfs(int x,int c){
    color[x] = c;v[c].pb(x);
    int len = g[x].size();
    rep1(j,0,len-1){
        int y = g[x][j];
        if (color[y]==-1) dfs(y,1-c);
    }
}

bool hungary(int x){
    int len = g[x].size();
    rep1(i,0,len-1){
        int y = g[x][i];
        if (!Try[y]){
            Try[y] = 1;
            if ( pre[y]==-1 || hungary(pre[y])){
                pre[y] = x;
                return true;
            }
        }
    }
    return false;
}

int main(){
    //Open();
    //Close();
    while (~scanf("%d",&n)){
        rep1(i,1,n) g[i].clear();
        rep1(i,1,n){
            int x,cnt;
            scanf("%d: ",&x);
            x++;
            scanf("(%d)",&cnt);
            rep1(j,1,cnt){
                int y;
                ri(y);
                y++;
                g[x].pb(y),g[y].pb(x);
            }
        }

        rep1(i,0,1) v[i].clear();
        ms(color,255);
        rep1(i,1,n)
            if (color[i]==-1)
                dfs(i,0);

        int len = v[0].size(),ans = 0;
        ms(pre,255);
        rep1(i,0,len-1){
            ms(Try,0);
            if (hungary(v[0][i])) ans++;
        }
        oi(n-ans);puts("");
    }
    return 0;
}