ural-1099-Work Scheduling(裸带花树)

题意:

有N个人,有限对的人可以在一起工作,问最多能有多少对.

分析:

任意图的最大匹配

// File MAXName: 1099.cpp
// Author: Zlbing
// Created Time: 2013/8/31 14:37:38

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
#define MAXN 250
#define SET(a,b) memset(a,b,sizeof(a))
deque<int> Q;
//g[i][j]存放关系图：i,j是否有边,match[i]存放i所匹配的点
//建图开始初始化g
//最终匹配方案为match
//复杂度O(n^3)
//点是从1到n的 
bool g[MAXN][MAXN],inque[MAXN],inblossom[MAXN],inpath[MAXN];
int match[MAXN],pre[MAXN],base[MAXN];

//找公共祖先
int findancestor(int u,int v)
{
    memset(inpath,false,sizeof(inpath));
    while(1)
    {
        u=base[u];
        inpath[u]=true;
        if(match[u]==-1)break;
        u=pre[match[u]];
    }
    while(1)
    {
        v=base[v];
        if(inpath[v])return v;
        v=pre[match[v]];
    }
}

//压缩花
void reset(int u,int anc)
{
    while(u!=anc)
    {
        int v=match[u];
        inblossom[base[u]]=1;
        inblossom[base[v]]=1;
        v=pre[v];
        if(base[v]!=anc)pre[v]=match[u];
        u=v;
    }
}

void contract(int u,int v,int n)
{
    int anc=findancestor(u,v);
    SET(inblossom,0);
    reset(u,anc);reset(v,anc);
    if(base[u]!=anc)pre[u]=v;
    if(base[v]!=anc)pre[v]=u;
    for(int i=1;i<=n;i++)
        if(inblossom[base[i]])
        {
            base[i]=anc;
            if(!inque[i])
            {
                Q.push_back(i);
                inque[i]=1;
            }
        }
}

bool bfs(int S,int n)
{
    for(int i=0;i<=n;i++)pre[i]=-1,inque[i]=0,base[i]=i;
    Q.clear();Q.push_back(S);inque[S]=1;
    while(!Q.empty())
    {
        int u=Q.front();Q.pop_front();
        for(int v=1;v<=n;v++)
        {
            if(g[u][v]&&base[v]!=base[u]&&match[u]!=v)
            {
                if(v==S||(match[v]!=-1&&pre[match[v]]!=-1))contract(u,v,n);
                else if(pre[v]==-1)
                {
                    pre[v]=u;
                    if(match[v]!=-1)Q.push_back(match[v]),inque[match[v]]=1;
                    else
                    {
                        u=v;
                        while(u!=-1)
                        {
                            v=pre[u];
                            int w=match[v];
                            match[u]=v;
                            match[v]=u;
                            u=w;
                        }
                        return true;
                    }
                }
            }
        }
    }
    return false;
}

int solve(int n)
{
    SET(match,-1);
    int ans=0;
    for(int i=1;i<=n;i++)
        if(match[i]==-1&&dfs(i,n))
            ans++;
    return ans;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int a,b;
        while(~scanf("%d%d",&a,&b))
        {
            g[a][b]=g[b][a]=1;
        }
        int ans=solve(n);
        printf("%d
",ans*2);
        for(int i=1;i<=n;i++)
        {
            if(match[i]!=-1)
            {
                printf("%d %d
",i,match[i]);
                match[match[i]]=-1;
                match[i]=-1;
            }
        }
    }
    return 0;
}