Swap
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2514 Accepted Submission(s): 900
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
题目大意：给你一个 n*n的 矩阵，使用行交换或者列交换的方式来使得矩阵的主对角线（左上到右下）都是1.
分析：对于这个题，不需要最少的操作次数，那么我们无论是怎样做，只要是搞定了这个问题，我们就胜利了，对于输出-1的情况，我们来行列匹配看看能否使得所有行都有列来匹配就好，如果没有，那么就不是完美匹配，也就是说输出-1，否则就一定有结果输出。
思路：
完美匹配：如果一个图的某个匹配中，所有的顶点都是匹配点，那么它就是一个完美匹配。图
4 是一个完美匹配。显然，完美匹配一定是最大匹配（完美匹配的任何一个点都已经匹配，添加一条新的匹配边一定会与已有的匹配边冲突）。但并非每个图都存在完美匹配。
遍历所有行，如果都有列匹配的话，就是一个完美匹配，否则输出-1.
如果是完美匹配，那么交换pri【】数组，使得pri【i】=i、
AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 505;
int map[maxn][maxn]; //图的数组.
int vis[maxn]; //标记数组.
int pri[maxn];
int ans[maxn*2][2];
int n;
int find(int x)
{
for(int i=1; i<=n; i++)
{
if(vis[i]==0&&map[i][x])
{
vis[i]=1;
if(pri[i]==-1||find(pri[i]))
{
pri[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
while(~scanf("%d",&n))
{
memset(pri,-1,sizeof(pri));
for(int i=1; i<=n; i++)
{
for(int j=1; j<=n; j++)
{
scanf("%d",&map[i][j]);
}
}
int ok=1;
for(int i=1; i<=n; i++)
{
memset(vis,0,sizeof(vis));
if(find(i))continue;
else ok=0;
}
if(ok==0){printf("-1\n");continue;}
int cont=0;
for(int i=1;i<=n;i++)
{
if(pri[i]==i)continue;
for(int j=1;j<=n;j++)
{
if(pri[j]==i)
{
swap(pri[i],pri[j]);
ans[cont][0]=i;ans[cont++][1]=j;
}
}
}
printf("%d\n",cont);
for(int i=0;i<cont;i++)
{
printf("R %d %d\n",ans[i][0],ans[i][1]);
}
}
}