Ural1109_Conference(二分图最大匹配/匈牙利算法/网络最大流)

解题报告

二分图第一题。

题目描写叙述：

为了參加即将召开的会议，A国派出M位代表，B国派出N位代表，（N，M<=1000）

会议召开前，选出K队代表，每对代表必须一个是A国的，一个是B国的;

要求每个代表要与还有一方的一个代表联系，除了能够直接联系，也能够电话联系，求电话联系最少

思路：

电话联系最少就要使直接联系最大，又是一一匹配关系，就是二分图的最大匹配。

以下是匈牙利算法。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 1050
#define M 1050
using namespace std;
int mmap[M][N],vis[N],pre[N],n,m,k;
int dfs(int x)
{
    for(int i=1;i<=n;i++)
    {
        if(!vis[i]&&mmap[x][i])
        {
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i]))
            {
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,u,v;
    memset(pre,-1,sizeof(pre));
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&m,&n,&k);
    for(i=0;i<k;i++)
    {
        scanf("%d%d",&u,&v);
        mmap[u][v]=1;
    }
    int ans=0;
    for(i=1;i<=m;i++)
    {
        memset(vis,0,sizeof(vis));
        ans+=dfs(i);
    }
    printf("%d\n",n+m-ans);
}

二分最大匹配也能够用最大流做，当试试模板

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define inf 99999999
#define N 1050
#define M 1050
using namespace std;
int mmap[M+N][N+M],vis[N],l[N+M],n,m,k;
int bfs()
{
    queue<int>Q;
    Q.push(0);
    memset(l,-1,sizeof(l));
    l[0]=0;
    while(!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        for(int i=0;i<=n+m+1;i++)
        {
            if(l[i]==-1&&mmap[u][i])
            {
                l[i]=l[u]+1;
                Q.push(i);
            }
        }
    }
    if(l[n+m+1]>0)
    return 1;
    return 0;
}
int dfs(int x,int f)
{
    int a,i;
    if(x==n+m+1)
    return f;
    for(i=0;i<=n+m+1;i++)
    {
        if(mmap[x][i]&&l[i]==l[x]+1&&(a=dfs(i,min(f,mmap[x][i]))))
        {
            mmap[x][i]-=a;
            mmap[i][x]+=a;
            return a;
        }
    }
    return 0;
}
int main()
{
    int i,j,u,v;
    memset(vis,0,sizeof(vis));
    scanf("%d%d%d",&m,&n,&k);
    for(i=1;i<=m;i++)
    mmap[0][i]=1;
    for(i=m+1;i<=m+n;i++)
    mmap[i][m+n+1]=1;
    for(i=0;i<k;i++)
    {
        scanf("%d%d",&u,&v);
        mmap[u][v+m]=1;
    }
    int ans=0,a;
    while(bfs())
        while(a=dfs(0,inf))
        ans+=a;
    printf("%d\n",n+m-ans);
}

Conference

Time limit: 0.5 second
Memory limit: 64 MB

On the upcoming conference were sent M representatives of country A and N representatives of country B (M and N ≤ 1000). The representatives were identified with 1, 2, …, M for country A and 1, 2, …, N for country B. Before the conference K pairs of representatives were chosen. Every such pair consists of one member of delegation A and one of delegation B. If there exists a pair in which both member #i of A and member #j of B are included then #i and #j can negotiate. Everyone attending the conference was included in at least one pair. The CEO of the congress center wants to build direct telephone connections between the rooms of the delegates, so that everyone is connected with at least one representative of the other side, and every connection is made between people that can negotiate. The CEO also wants to minimize the amount of telephone connections. Write a program which given M, N, K and K pairs of representatives, finds the minimum number of needed connections.

Input

The first line of the input contains M, N and K. The following K lines contain the choosen pairs in the form of two integers p1and p2, p1 is member of A and p2 is member of B.

Output

The output should contain the minimum number of needed telephone connections.

Sample

input	output
3 2 4 1 1 2 1 3 1 3 2	3

Problem Source: Bulgarian National Olympiad Day #1