HDU2819(KB10-E 二分图最大匹配)

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3800    Accepted Submission(s): 1401
Special Judge

Problem Description

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?

Input

There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.

Output

For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”.

Sample Input

2
0 1
1 0
2
1 0
1 0

Sample Output

1
R 1 2
-1

Source

2009 Multi-University Training Contest 1 - Host by TJU

按行交换。对行进行1-n编号，对列也进行编号。

若第i行的第j1、j2……列有1，则节点i与j1、j2……连边。

二分图跑最大匹配，为完美匹配方可。

matching数组保存每个点的匹配，模拟一下交换输出答案。

 //2017-08-26
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 using namespace std;

 const int N = ;
 const int M = ;
 int head[N], tot;
 struct Edge{
     int to, next;
 }edge[M];

 void init(){
     tot = ;
     memset(head, -, sizeof(head));
 }

 void add_edge(int u, int v){
     edge[tot].to = v;
     edge[tot].next = head[u];
     head[u] = tot++;

     edge[tot].to = u;
     edge[tot].next = head[v];
     head[v] = tot++;
 }

 int n;
 int matching[N];
 int check[N];
 bool dfs(int u){
     for(int i =  head[u]; i != -; i = edge[i].next){
         int v = edge[i].to;
         if(!check[v]){//要求不在交替路
             check[v] = ;//放入交替路
             if(matching[v] == - || dfs(matching[v])){
                 //如果是未匹配点，说明交替路为增广路，则交换路径，并返回成功
                 matching[u] = v;
                 matching[v] = u;
                 return true;
             }
         }
     }
     return false;//不存在增广路
 }

 //hungarian: 二分图最大匹配匈牙利算法
 //input: null
 //output: ans 最大匹配数
 int hungarian(){
     int ans = ;
     memset(matching, -, sizeof(matching));
     for(int u = ; u <= n; u++){
         if(matching[u] == -){
             memset(check, , sizeof(check));
             if(dfs(u))
               ans++;
         }
     }
     return ans;
 }

 const int MAXID = ;
 int a[N], b[N], cnt;

 int main()
 {
     std::ios::sync_with_stdio(false);
     //freopen("inputE.txt", "r", stdin);
     while(cin>>n){
         init();
         int v;
         for(int i = ; i <= n; i++){
             for(int j = ; j <= n; j++){
                 cin>>v;
                 if(v){
                     add_edge(i, j+MAXID);
                 }
             }
         }
         int match = hungarian();
         if(match != n)cout<<-<<endl;
         else{
             for(int i = ; i <= n; i++)
                   matching[i]-=;
             cnt = ;
             for(int i = ; i <= n; i++){
                 for(int j = ; j <= n; j++){
                     if(i == j)continue;
                     if(matching[j] == i){
                         swap(matching[i], matching[j]);
                         a[cnt] = i;
                         b[cnt++] = j;
                     }
                 }
             }
             cout<<cnt<<endl;
             for(int i = ; i < cnt; i++)
                   cout<<"R "<<a[i]<<" "<<b[i]<<endl;
         }
     }

     return ;
 }