P3386 【模板】二分图匹配 Ek 与 dinic

题目背景

二分图

题目描述

给定一个二分图,结点个数分别为n,m,边数为e,求二分图最大匹配数

输入输出格式

输入格式:

第一行,n,m,e

第二至e+1行,每行两个正整数u,v,表示u,v有一条连边

输出格式:

共一行,二分图最大匹配

输入输出样例

输入样例#1: 复制
1 1 1
1 1
输出样例#1: 复制
1

说明  n,m10001≤u≤n 1vm

因为数据有坑,可能会遇到 v>mv>mv>m 的情况。请把 v>mv>mv>m 的数据自觉过滤掉。

算法:二分图匹配

通过引一个源点 到 n 的所有点和m的所有点到一个汇点,将二分图转化,这个新图的最大流(即不存在增广路时)就等于二分图的最大匹配。

这里m没给定,但显然是略大的(>10000)所以用O(V*E*E)的EK算法超时了....

这里是EK算法的代码

#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 1000;
const int F = 2010;
const int INF = 1E9;
typedef long long LL;
typedef double DB;
inline int get(){
	char c;
	while((c = getchar()) < '0' || c > '9');
	int cnt = c - '0';
	while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
	return cnt;
}
int N,M,E;
struct edge{
	int fr,to,f;
	edge(int u,int v,int w):fr(u),to(v),f(w){}
};
vector e;
vector g[2 * MAXN + 20];
int a[2 * MAXN + 20];
int pa[2 * MAXN + 20];
int maxf = 0;
inline void zdl(int s,int t){
	while(1){
		memset(a,0,sizeof(a));
		a[s] = INF;
		queue q;
		q.push(s);
		while(!q.empty()){
			int f = q.front(); q.pop();
			for(int i = 0; i < g[f].size(); i ++){
				edge k = e[g[f][i]];
				if(!a[k.to] && k.f > 0){
					a[k.to] = 1;
					pa[k.to] = g[f][i];
					q.push(k.to);
				}
				if(a[t]) break;
			}
			if(a[t]) break;
		}
		if(!a[t]) break;
		for(int i = t; i != s; i = e[pa[i]].fr){
			e[pa[i]].f -= 1;
			e[pa[i] ^ 1].f += 1;
		}
		maxf += a[t];
	}
	return;
}
int main(){
	#ifdef lwy
		freopen("1.txt","r",stdin);
/*	#else
		freopen(".in","r",stdin);
		freopen(".out","w",stdout);*/ 
	#endif
	N = get(); M = get(); E = get();
	for(int i = 1; i <= E; i ++){
		int u,v;
		u = get(); v = get();
		if(u > N || v > M) continue;
		v += MAXN;
		e.push_back(edge(u,v,1));
		e.push_back(edge(v,u,0));
		int m = e.size();
		g[v].push_back(m - 1);
		g[u].push_back(m - 2);
	} 
	for(int i = 1; i <= N; i ++){
		e.push_back(edge(0,i,1));
		e.push_back(edge(i,0,0));
		int m = e.size();
		g[i].push_back(m - 1);
		g[0].push_back(m - 2);
	}
	for(int i = 1; i <= M; i++){
		e.push_back(edge(i + MAXN,F,1));
		e.push_back(edge(F,i + MAXN,0));
		int m = e.size();
		g[F].push_back(m - 1);
		g[i + MAXN].push_back(m - 2);
	}
	zdl(0,F);
	printf("%d",maxf);
	return 0;
}

由于二分图需要引一个超级源点和一个超级汇点,所以这个图有可能是稠密图,这样Ek算法的O(V * E *E)就显得非常吃力了....

所以要用dinic算法 

可以证明dinic算法在二分图的复杂度是O(根号V * E),这样可以轻松跑过大的点了。

#include
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 2000;
const int INF = 1E9;
typedef long long LL;
typedef double DB;
inline int get(){
    char c;
    while((c = getchar()) < '0' || c > '9');
    int cnt = c - '0';
    while((c = getchar()) >= '0' && c <= '9') cnt = cnt * 10 + c - '0';
    return cnt;
}
int N,M,E,S,T;
int maxf = 0;
struct edge{
    int fr,to,fl;
    edge(int u,int v,int w): fr(u),to(v),fl(w){}
};
vector e;
vector g[MAXN + 10];
int dep[MAXN + 10];
bool vis[MAXN + 10];
int cur[MAXN + 10];
inline bool bfs(){
    memset(vis,false,sizeof(vis));
    memset(dep,0,sizeof(dep));
    vis[S] = true; 
    queue q;
    q.push(S); 
    while(!q.empty()){
        int f = q.front(); q.pop();
        for(int i = 0; i < g[f].size(); i ++){
            edge k = e[g[f][i]];
            if(!vis[k.to] && k.fl > 0){ // k.fl
                dep[k.to] = dep[f] + 1;
                vis[k.to] = true;
                q.push(k.to);
            }
        }
    }
    return vis[T];
}
inline int dfs(int x,int a){
    if(x == T || a == 0) return a;
    int flowt = 0,flown; // flowt = 0
    for(int& i = cur[x]; i < g[x].size(); i ++){
        edge k = e[g[x][i]];
        if(dep[k.to] == dep[x] + 1){
            int flown = dfs(k.to,min(a,k.fl));
            if(flown > 0){
                e[g[x][i]].fl -= flown;
                e[g[x][i] ^ 1].fl += flown;
                flowt += flown;
                a -= flown;
                if(a == 0) break;
            }
        }
    }
    return flowt;
}
int main(){
    #ifdef lwy
        freopen("1.txt","r",stdin);
/*	#else
        freopen(".in","r",stdin);
        freopen(".out","w",stdout);*/ 
    #endif
    N = get(); M = get(); E = get();
    S = 0; T = N + M + 1;
    for(int i = 1; i <= E; i ++){
        int u,v;
        u = get(); v = get(); v += N;
        if(u > N || v > M + N) continue;
        e.push_back(edge(u,v,1));
        e.push_back(edge(v,u,0));
        int m = e.size();
        g[v].push_back(m - 1);
        g[u].push_back(m - 2);
    }
    for(int i = 1; i <= N; i ++){
        e.push_back(edge(0,i,1));
        e.push_back(edge(i,0,0));
        int m = e.size();
        g[i].push_back(m - 1);
        g[0].push_back(m - 2);
    }
    for(int i = N + 1; i <= N + M; i ++){
        e.push_back(edge(i,T,1));
        e.push_back(edge(T,i,0));
        int m = e.size();
        g[T].push_back(m - 1);
        g[i].push_back(m - 2);
    }
    while(bfs()){
        memset(cur,0,sizeof(cur));
        maxf += dfs(S,INF);
    }
    printf("%d",maxf);
    return 0;
}





你可能感兴趣的:(网络流,图论)