HDU2819 Swap —— 二分图最大匹配

题目链接：https://vjudge.net/problem/HDU-2819

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4003    Accepted Submission(s): 1478
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

 

Recommend

gaojie

题解：

题意：给出一个大小为n*n的01矩阵，问能否通过交换行或列，使得主对角线上的数全为1？

1.用匈牙利算法求出最大匹配数cnt，如果cnt等于n，则可以实现。

2.可知，我们可以通过只交换行或者只交换列，就能使得主对角线上的数全为1。

3.枚举每一行i，对于第i行，找到与第i列匹配的那一行k，然后交换第i行和第k行。这样就满足了第i行有1在对角线上。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <vector>
 #include <map>
 #include <set>
 #include <queue>
 #include <sstream>
 #include <algorithm>
 using namespace std;
 const int INF = 2e9;
 const int MOD = 1e9+;
 const int MAXN = +;

 int n;
 int M[MAXN][MAXN], id[MAXN][MAXN], xlink[MAXN], ylink[MAXN];
 bool vis[MAXN];

 bool dfs(int u)
 {
     for(int i = ; i<=n; i++)
     if(M[u][i] && !vis[i])
     {
         vis[i] = true;
         if(ylink[i]==- || dfs(ylink[i]))
         {
             ylink[i] = u;
             xlink[u] = i;
             return true;
         }
     }
     return false;
 }

 int hungary()
 {
     int ret = ;
     memset(xlink, -, sizeof(xlink));
     memset(ylink, -, sizeof(ylink));
     for(int i = ; i<=n; i++)
     {
         memset(vis, , sizeof(vis));
         if(dfs(i)) ret++;
     }
     return ret;
 }

 int main()
 {
     while(scanf("%d", &n)!=EOF)
     {
         for(int i = ; i<=n; i++)
         for(int j = ; j<=n; j++)
             scanf("%d", &M[i][j]);

         int cnt = hungary();
         if(cnt<n)
         {
             printf("-1\n");
             continue;
         }

         int ans = , op[MAXN][];
         for(int i = ; i<n; i++)
         for(int j = i+; j<=n; j++)
         {
             if(xlink[j]==i)
             {
                 swap(xlink[j], xlink[i]);
                 op[++ans][] = i; op[ans][] = j;
                 break;
             }
         }

         printf("%d\n", ans);
         for(int i = ; i<=ans; i++)
             printf("R %d %d\n", op[i][], op[i][]);
     }
 }