ural 1099. Work Scheduling 一般圖最大匹配

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;//點數(1->n)
int head;
int tail;
int Start;
int Finish;
int link[N];     //表示哪個點匹配了哪個點
int Father[N];   //這個就是增廣路的Father……但是用起來太精髓了
int Base[N];     //該點屬於哪朵花
int Q[N];
bool mark[N];
bool mat[N][N];//鄰接矩陣
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;} }  //尋找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]; //對於BFS樹中的父邊是匹配邊的點，Father向後跳
        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;     //注意不能從lca這個奇環的關鍵點跳回來
    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;     //要注意如果本來連向BFS樹中父結點的邊是非匹配邊的點，可能是沒有入隊的
          }
      }
}


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)   //無意義的邊
            if ( Start==y || link[y] && Father[link[y]] )    //精髓地用Father表示該點是否
                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;//一般圖最大匹配  最大點數
    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;
}