【hihocoder 1122】二分图二•二分图最大匹配之匈牙利算法

【Link】:https://hihocoder.com/problemset/problem/1122

【Description】

【Solution】

二分图匹配,匈牙利算法模板题;
这里我先把染成0的放在一个vector里面,然后再进行匈牙利算法.

【NumberOf WA】

【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;

vector <int> g[N+100],v[2];
int pre[N+10],Try[N+10],color[N+10];
int n,m;

bool hungary(int x){
    int len = g[x].size();
    for (int i = 0;i <= len-1;i++){
        int y = g[x][i];
        if (!Try[y]){
            Try[y] = 1;
            if ( pre[y] == -1 || hungary(pre[y])){
                pre[y] = x;
                return 1;
            }
        }
    }
    return 0;
}

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

int main(){
    //Open();
    //Close();
    ri(n),ri(m);
    rep1(i,1,m){
        int x,y;
        ri(x),ri(y);
        g[x].pb(y),g[y].pb(x);
    }

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

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

    oi(ans);puts("");
    return 0;
}