POJ 1719 二分图最大匹配（记录路径）

Shooting Contest

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4097		Accepted: 1499		Special Judge

Description

Welcome to the Annual Byteland Shooting Contest. Each competitor will shoot to a target which is a rectangular grid. The target consists of r*c squares located in r rows and c columns. The squares are coloured white or black. There are exactly two white squares and r-2 black squares in each column. Rows are consecutively labelled 1,..,r from top to bottom and columns are labelled 1,..,c from left to right. The shooter has c shots.

A volley of c shots is correct if exactly one white square is hit in each column and there is no row without white square being hit. Help the shooter to find a correct volley of hits if such a volley exists. 
Example 
Consider the following target: 

Volley of hits at white squares in rows 2, 3, 1, 4 in consecutive columns 1, 2, 3, 4 is correct. 
Write a program that: verifies whether any correct volley of hits exists and if so, finds one of them.

Input

The first line of the input contains the number of data blocks x, 1 <= x <= 5. The following lines constitute x blocks. The first block starts in the second line of the input file; each next block starts directly after the previous one.

The first line of each block contains two integers r and c separated by a single space, 2 <= r <= c <= 1000. These are the numbers of rows and columns, respectively. Each of the next c lines in the block contains two integers separated by a single space. The integers in the input line i + 1 in the block, 1 <= i <= c, are labels of rows with white squares in the i-th column.

Output

For the i-th block, 1 <= i <= x, your program should write to the i-th line of the standard output either a sequence of c row labels (separated by single spaces) forming a correct volley of hits at white squares in consecutive columns 1, 2, ..., c, or one word NO if such a volley does not exists.

Sample Input

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

Sample Output

2 3 1 4
NO

Source

CEOI 1997

题目意思：

有n*m的格子，每个格子要么是白色要么是黑色。题目给定每列有2个白色的格子，其他的格子为黑色。每列选出一个白色的格子，使得每一行至少有一个白色格子被选出，若能满足条件则输出每列选出白色格子的行数，否则输出NO。

思路：

列标号放在左边，行标号放在右边，若该列a可选择第b行和第c行，则a-b连边，a-c连边。

在增广过程中记录to[i]和from[j]即第i列选择的行的标号和第j行被选择的列的标号。

init：to 0;from -1

1、若n>m，一定不可满足条件

2、若增广结束后存在from[i]=-1，则第i行一定不能被选择，输出NO

3、若增广结束后存在to[i]=0，那么第i列选择a和b其中一个即可。

代码：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <vector>
 #include <queue>
 #include <cmath>
 #include <set>
 using namespace std;

 #define N 1005
 #define inf 999999999

 int max(int x,int y){return x>y?x:y;}
 int min(int x,int y){return x<y?x:y;}
 int abs(int x,int y){return x<?-x:x;}

 int from[N];
 int to[N];
 bool visited[N];
 int n, m;
 vector<int>ve[N];

 int march(int u){
     int i, v;
     for(i=;i<ve[u].size();i++){
         v=ve[u][i];
         if(!visited[v]){
             visited[v]=true;
             if(from[v]==-||march(from[v])){
                 from[v]=u;
                 to[u]=v;
                 return ;
             }
         }
     }
     return ;
 }

 main()
 {
     int t, i, j, k, u, v;
     cin>>t;
     while(t--){
         scanf("%d %d",&n,&m);
         for(i=;i<=m;i++) ve[i].clear();
         for(i=;i<=m;i++){
             scanf("%d %d",&u,&v);
             ve[i].push_back(u);
             ve[i].push_back(v);
         }
         if(n>m) {
             printf("NO\n");continue;
         }
         memset(from,-,sizeof(from));
         int num=;
         for(i=;i<=m;i++){
             memset(visited,false,sizeof(visited));
             march(i);
         }
         int f=;
         for(i=;i<=n;i++){
             if(from[i]==-){
                 f=;break;
             }
         }
         if(!f){
             printf("NO\n");continue;
         }
         for(i=;i<=m;i++){
             if(!to[i]){
                 to[i]=ve[i][];
             }
         }
         printf("%d",to[]);
         for(i=;i<=m;i++) printf(" %d",to[i]);
         cout<<endl;
     }
 }