hdu 2819 Swap（二分图最大匹配，输出路径）

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2712    Accepted Submission(s): 971
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

题意：给一个矩阵，问能否通过交换行和列使得主对角线都是1

思路：可以参考这位大神，写得很详细了。

点击打开链接

最后输出路径我们需要模拟这个交换的过程，比如把第一行跟第三行交换的话，我们记录第一次，然后模拟这个交换的过程

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 210
int ma

;
int line
,vis
;
int n,a
,b
;
int can(int t)
{
for(int i=1; i<=2*n; i++)
{
if(!vis[i]&&ma[t][i])
{
vis[i]=1;
if(line[i]==-1||can(line[i]))
{
line[i]=t;
return 1;
}
}
}
return 0;
}
int Maxmatch()
{
int ans=0;
for(int i=1; i<=n; i++)
{
memset(vis,0,sizeof(vis));
if(can(i)) ans++;
else break;
}
return ans;
}
int main()
{
while(~scanf("%d",&n))
{
memset(line,-1,sizeof(line));
memset(ma,0,sizeof(ma));
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
scanf("%d",&ma[i][j]);
int ans=Maxmatch();
if(ans!=n) printf("-1\n");
else
{
int num=0;
for(int i=1;i<=n;i++)///模拟交换的过程
{
int k;
for(k=1;k<=n;k++)
if(line[k]==i)
break;
if(i!=k)
{
a[num]=i,b[num++]=k;
swap(line[i],line[k]);///交换第i列跟第k列
}
}
printf("%d\n",num);
for(int i=0;i<num;i++)
printf("C %d %d\n",a[i],b[i]);
}
}
return 0;
}