ural 1099. Work Scheduling 一般图最大匹配

There is certain amount of night guards that are available to protect the local junkyard from possible junk robberies. These guards need to scheduled in pairs, so that each pair guards at different night. The junkyard CEO ordered you to write a program which given the guards characteristics determines the maximum amount of scheduled guards (the rest will be fired). Please note that each guard can be scheduled with only one of his colleagues and no guard can work alone.

Input

The first line of the input contains one number N ≤ 222 which is the amount of night guards. Unlimited number of lines consisting of unordered pairs (i, j) follow, each such pair means that guard #i and guard #j can work together, because it is possible to find uniforms that suit both of them (The junkyard uses different parts of uniforms for different guards i.e. helmets, pants, jackets. It is impossible to put small helmet on a guard with a big head or big shoes on guard with small feet). The input ends with Eof.

Output

You should output one possible optimal assignment. On the first line of the output write the even number C, the amount of scheduled guards. Then output C/2 lines, each containing 2 integers (i, j) that denote that i and j will work together.

Sample

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

//

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=250;
int n;//点数(1->n)
int head;
int tail;
int Start;
int Finish;
int link[N];     //表示哪个点匹配了哪个点
int Father[N];   //这个就是增广路的Father……但是用起来太精髓了
int Base[N];     //该点属于哪朵花
int Q[N];
bool mark[N];
bool mat[N][N];//邻接矩阵
bool InBlossom[N];
bool in_Queue[N];




void BlossomContract(int x,int y){
    memset(mark,0,sizeof(mark));
    memset(InBlossom,0,sizeof(InBlossom));
    #define pre Father[link[i]]
    int lca,i;
    for (i=x;i;i=pre) {i=Base[i]; mark[i]=true; }
    for (i=y;i;i=pre) {i=Base[i]; if (mark[i]) {lca=i; break;} }  //寻找lca之旅……一定要注意i=Base[i]
    for (i=x;Base[i]!=lca;i=pre){
        if (Base[pre]!=lca) Father[pre]=link[i]; //对于BFS树中的父边是匹配边的点，Father向后跳
        InBlossom[Base[i]]=true;
        InBlossom[Base[link[i]]]=true;
    }
    for (i=y;Base[i]!=lca;i=pre){
        if (Base[pre]!=lca) Father[pre]=link[i]; //同理
        InBlossom[Base[i]]=true;
        InBlossom[Base[link[i]]]=true;
    }
    #undef pre
    if (Base[x]!=lca) Father[x]=y;     //注意不能从lca这个奇环的关键点跳回来
    if (Base[y]!=lca) Father[y]=x;
    for (i=1;i<=n;i++)
      if (InBlossom[Base[i]]){
          Base[i]=lca;
          if (!in_Queue[i]){
              Q[++tail]=i;
              in_Queue[i]=true;     //要注意如果本来连向BFS树中父结点的边是非匹配边的点，可能是没有入队的
          }
      }
}


void Change(){
    int x,y,z;
    z=Finish;
    while (z){
        y=Father[z];
        x=link[y];
        link[y]=z;
        link[z]=y;
        z=x;
    }
}


void FindAugmentPath(){
    memset(Father,0,sizeof(Father));
    memset(in_Queue,0,sizeof(in_Queue));
    for (int i=1;i<=n;i++) Base[i]=i;
    head=0; tail=1;
    Q[1]=Start;
    in_Queue[Start]=1;
    while (head!=tail){
        int x=Q[++head];
        for (int y=1;y<=n;y++)
          if (mat[x][y] && Base[x]!=Base[y] && link[x]!=y)   //无意义的边
            if ( Start==y || link[y] && Father[link[y]] )    //精髓地用Father表示该点是否
                BlossomContract(x,y);
            else if (!Father[y]){
                Father[y]=x;
                if (link[y]){
                    Q[++tail]=link[y];
                    in_Queue[link[y]]=true;
                }
                else{
                    Finish=y;
                    Change();
                    return;
                }
            }
    }
}


void Edmonds(){
    memset(link,0,sizeof(link));
    for (Start=1;Start<=n;Start++)
      if (link[Start]==0)
        FindAugmentPath();
}


void output(){
    memset(mark,0,sizeof(mark));
    int cnt=0;//一般图最大匹配  最大点数
    for (int i=1;i<=n;i++)
      if (link[i]) cnt++;
    printf("%d\n",cnt);
    for (int i=1;i<=n;i++)
      if (!mark[i] && link[i]){
          mark[i]=true;//i和link[i]匹配
          mark[link[i]]=true;
          printf("%d %d\n",i,link[i]);
      }
}


int main(){
    int x,y;
    scanf("%d",&n);
    memset(mat,0,sizeof(mat));
    while (scanf("%d%d",&x,&y)!=EOF)
      mat[x][y]=mat[y][x]=1;
    Edmonds();
    output();
    return 0;
}