HDU-4687 Boke and Tsukkomi

一般图匹配问题，带花树算法

要求给出的边中那些是冗余的，就是不再任何一个最大匹配里

把每条边的点都去掉，检查匹配值是否小于最大匹配值-1，如果是的话就是冗余边

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
const int N=250;
int n;
bool gap[N][N];
int match[N];
bool inque[N],inpath[N],inblo[N];
int head,tail;
int que[N];
int start,finish;
int newbase;
int father[N],base[N];
int cnt;
void push(int u)
{
    que[tail]=u;
    tail++;
    inque[u]=true;
}
int pop()
{
    int res=que[head];
    head++;
    return res;
}
int FCA(int u,int v)
{
    memset(inpath,false,sizeof(inpath));
    while(true)
    {
        u=base[u];
        inpath[u]=true;
        if(u==start) break;
        u=father[match[u]];
    }
    while(true)
    {
        v=base[v];
        if(inpath[v]) break;
        v=father[match[v]];
    }
    return v;
}
void RT(int u)
{
    int v;
    while(base[u]!=newbase)
    {
        v=match[u];
        inblo[base[u]]=inblo[base[v]]=true;
        u=father[v];
        if(base[u]!=newbase) father[u]=v;
    }
}
void BC(int u,int v)
{
    newbase=FCA(u,v);
    memset(inblo,false,sizeof(inblo));
    RT(u);
    RT(v);
    if(base[u]!=newbase) father[u]=v;
    if(base[v]!=newbase) father[v]=u;
    for(int tu=1;tu<=n;tu++)
        if(inblo[base[tu]])
        {
            base[tu]=newbase;
            if(!inque[tu]) push(tu);
        }
}
void FAP()
{
    memset(inque,false,sizeof(inque));
    memset(father,0,sizeof(father));
    for(int i=1;i<=n;i++)
        base[i]=i;
    head=tail=1;
    push(start);
    finish=0;
    while(head<tail)
    {
        int u=pop();
        for(int v=1;v<=n;v++)
            if(gap[u][v]&&(base[u]!=base[v])&&(match[u]!=v))
            {
                if((v==start)||((match[v]>0)&&father[match[v]]>0))
                    BC(u,v);
                else if(father[v]==0)
                {
                    father[v]=u;
                    if(match[v]>0)
                        push(match[v]);
                    else
                    {
                        finish=v;
                        return;
                    }
                }
            }
    }
}
void AP()
{
    int u,v,w;
    u=finish;
    while(u>0)
    {
        v=father[u];
        w=match[v];
        match[v]=u;
        match[u]=v;
        u=w;
    }
}
void ED()
{
    memset(match,0,sizeof(match));
    for(int u=1;u<=n;u++)
        if(match[u]==0)
        {
            start=u;
            FAP();
            if(finish>0) AP();
        }
}
int a[N],b[N];
int ans[N];
bool g[N][N];
int main()
{
    int u,v;
    int m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(gap,false,sizeof(gap));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a[i],&b[i]);
            gap[a[i]][b[i]]=gap[b[i]][a[i]]=true;
        } 
        ED();
        cnt=0;
        int num=0;
        for(int i=1;i<=n;i++)
            if(match[i]) num++;
        num/=2;
        for(int i=0;i<m;i++)
        {
            memcpy(g,gap,sizeof(gap));
            for(int j=1;j<=n;j++)
                gap[a[i]][j]=gap[j][a[i]]=gap[b[i]][j]=gap[j][b[i]]=false;
            ED();
            int t=0;
            for(int j=1;j<=n;j++)
                if(match[j]) t++;
            t/=2;
            if(t<num-1) ans[cnt++]=i+1;
            memcpy(gap,g,sizeof(gap));
        }
        printf("%d\n",cnt);
        for(int i=0;i<cnt-1;i++)
            printf("%d ",ans[i]);
        if(cnt)
        printf("%d",ans[cnt-1]);
        printf("\n");
    }
    return 0;
}