hdu 4687 Boke and Tsukkomi（一般图匹配 带花树算法）

Boke and Tsukkomi

Time Limit: 3000/3000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)

Total Submission(s): 739 Accepted Submission(s): 231



Problem Description

A new season of Touhou M-1 Grand Prix is approaching. Girls in Gensokyo cannot wait for participating it. Before the registration, they have to decide which combination they are going to compete as. Every girl in Gensokyo is both a boke (funny girl) and a tsukkomi
(straight girl). Every candidate combination is made up of two girls, a boke and a tsukkomi. A girl may belong to zero or more candidate combinations, but one can only register as a member of one formal combination. The host of Touhou M-1 Grand Prix hopes
that as many formal combinations as possible can participate in this year. Under these constraints, some candidate combinations are actually redundant as it\'s impossible to register it as a formal one as long as the number of formal combinations has to be
maximized. So they want to figure out these redundant combinations and stop considering about them.





Input

There are multiple test cases. Process to the End of File.

The first line of each test case contains two integers: 1 ≤ N ≤ 40 and 1 ≤ M ≤ 123, where N is the number of girls in Gensokyo, and M is the number of candidate combinations. The following M lines are M candidate combinations, one by each line.
Each combination is represented by two integers, the index of the boke girl 1 ≤ Bi ≤ N and the index of the tsukkomi girl 1 ≤ Ti ≤ N, where Bi != Ti.



Output

For each test case, output the number of redundant combinations in the first line. Then output the space-separated indexes of the redundant combinations in ascending order in the second line.



Sample Input

4 4
1 3
2 3
2 4
3 1
6 6
1 2
3 2
3 4
5 2
5 4
5 6

Sample Output

1
2
3
2 4 5

n个人 m个匹配 求出多余的匹配

首先用带花树匹配求出 最大匹配数mmg 然后在枚举每一个匹配 删除这个匹配 再求匹配数

如果不等于mmg-1 则这个匹配是多余的

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 200
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' | ch == '\n');
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return a;
}

void Print(int a)    //输出外挂
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}
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
bool g[MAXN][MAXN],inque[MAXN],inblossom[MAXN],inpath[MAXN];
int match[MAXN],pre[MAXN],base[MAXN];
int n,m,mmg;
vector<int> res;

struct node
{
    int u,v;
}point[MAXN];
//找公共祖先
int findancestor(int u,int v)
{
    MEM(inpath,0);
    while(1)
    {
        u=base[u];
        inpath[u]=1;
        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);
    MEM(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 dfs(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 1;
                    }
                }
            }
        }
    }
    return 0;
}

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

int main()
{
    //fread;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        MEM(g,0);
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            g[u][v]=g[v][u]=1;
            point[i].u=u; point[i].v=v;
        }
        mmg=solve();
        res.clear();
        for(int i=1;i<=m;i++)
        {
            int u=point[i].u;
            int v=point[i].v;
            MEM(g,0);
            for(int j=1;j<=m;j++)
            {
                if(i!=j)
                {
                    int x=point[j].u,y=point[j].v;
                    if(x==u||x==v||y==u||y==v)  continue;
                    g[x][y]=g[y][x]=1;
                }
            }
            int k=solve();
            if(k!=mmg-1)  res.push_back(i);
        }
        printf("%d\n",res.size());
        for(int i=0;i<res.size();i++)
        {
            if(i) printf(" ");
            printf("%d",res[i]);
        }
        printf("\n");
    }
    return 0;
}