P3386 【模板】二分图匹配（匈牙利&最大流）

 P3386 【模板】二分图匹配

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,m leq 1000n,m≤1000 ， 1 leq u leq n1≤u≤n ， 1 leq v leq m1≤v≤m

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

算法：二分图匹配

code

匈牙利

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>

using namespace std;

const int N = 1010;
int e[N][N],vis[N],resut[N];
int n,m,E;

inline int read() {
    int x = 0,f = 1;char ch = getchar();
    for (; ch<'0'||ch>'9'; ch = getchar())
        if (ch=='-') f = -1;
    for (; ch>='0'&&ch<='9'; ch = getchar())
        x = x*10+ch-'0';
    return x*f;
}

bool dfs(int u) {
    for (int v=1; v<=m; ++v) {
        if (e[u][v] && !vis[v]) {
            vis[v] = true;
            if (!resut[v] || dfs(resut[v])) {
                resut[v] = u;
                return true;
            }
        }
    }
    return false;
}

void xyl() {
    int ans = 0;
    for (int i=1; i<=n; ++i) {
        memset(vis,false,sizeof(vis));
        if (dfs(i)) ans++;
    }
    printf("%d",ans);
}

int main() {
    n = read(),m = read(),E = read();
    for (int i=1; i<=E; ++i) {
        int a = read(),b = read();
        if (a >n || b > m) continue;
        e[a][b] = 1;
    }
    xyl();
    return 0;
}

更新2018-06-03

#include<cstdio>
#include<cstring>
#include<cctype>

const int N = 1010;

struct Edge{
    int to,nxt;
    Edge() {}
    Edge(int a,int b) {to = a,nxt = b;}
}e[1000100];
int head[N],match[N],tot;
bool vis[N];
int n,m;

inline int read() {
    int x = 0,f = 1;char ch = getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-') f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x * f;
}
bool dfs(int u) {
    for (int i=head[u]; i; i=e[i].nxt) {
        int v = e[i].to;
        if (!vis[v]) {
            vis[v] = true;
            if (!match[v] || dfs(match[v])) {
                match[v] = u;
                return true;
            }
        }
    }
    return false;
}
int main () {
    n = read(),m = read();int E = read();
    for (int i=1; i<=E; ++i) {
        int u = read(),v = read();
        if (u > n || v > m) continue;
        e[++tot] = Edge(v,head[u]);
        head[u] = tot;
    }
    int ans = 0;
    for (int i=1; i<=n; ++i) {
        memset(vis,false,sizeof(vis));
        if (dfs(i)) ans++;
    }
    printf("%d",ans);
    return 0;
}

最大流

#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 2010;
const int INF = 1e9;
struct Edge{
    int to,nxt,c;
    Edge() {}
    Edge(int x,int y,int z) {to = x,c = y,nxt = z;}
}e[1000100];
int q[1000100],L,R,S,T,tot = 1;
int dis[N],cur[N],head[N];

inline char nc() {
    static char buf[100000],*p1 = buf,*p2 = buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2) ? EOF :*p1++;
}
inline int read() {
    int x = 0,f = 1;char ch=nc();
    for (; ch<'0'||ch>'9'; ch=nc()) if(ch=='-')f=-1;
    for (; ch>='0'&&ch<='9'; ch=nc()) x=x*10+ch-'0';
    return x*f;
}
void add_edge(int u,int v,int c) {
    e[++tot] = Edge(v,c,head[u]);head[u] = tot;
    e[++tot] = Edge(u,0,head[v]);head[v] = tot;
}
bool bfs() {
    for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1;
    L = 1,R = 0;
    q[++R] = S;dis[S] = 1;
    while (L <= R) {
        int u = q[L++];
        for (int i=head[u]; i; i=e[i].nxt) {
            int v = e[i].to;
            if (dis[v] == -1 && e[i].c > 0) {
                dis[v] = dis[u]+1;q[++R] = v;
                if (v==T) return true;
            }
        }
    }
    return false;
}
int dfs(int u,int flow) {
    if (u==T) return flow;
    int used = 0;
    for (int &i=cur[u]; i; i=e[i].nxt) {
        int v = e[i].to;
        if (dis[v] == dis[u] + 1 && e[i].c > 0) {
            int tmp = dfs(v,min(flow-used,e[i].c));
            if (tmp > 0) {
                e[i].c -= tmp;e[i^1].c += tmp;
                used += tmp;
                if (used == flow) break;
            }
        }
    }
    if (used != flow) dis[u] = -1;
    return used;
}
int dinic() {
    int ret = 0;
    while (bfs()) ret += dfs(S,INF);
    return ret;
}
int main() {
    int n = read(),m = read(),E = read();
    S = n+m+1;T = n+m+2;
    for (int i=1; i<=E; ++i) {
        int u = read(),v = read();
        if (u > n || v > m) continue;
        add_edge(u,v+n,1);
    }
    for (int i=1; i<=n; ++i) add_edge(S,i,1);
    for (int i=1; i<=m; ++i) add_edge(i+n,T,1);
    int ans = dinic();
    printf("%d",ans);
    return 0;
}