Swap
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2152 Accepted Submission(s): 764
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
R 1 2
-1
Source
题意：如果可以交换行列，问主对角线能不能全为1
分析：要想主对角线全为1很明显要有N个行列不想同的点就行了，可以用二分图匹配计算出来多能有几个。如果小与N就不能。输出要是对的就行，不必和答案一样
#include<iostream>
#include<cstdio>
#include<cstring> using namespace std; #define N 1100 int n, vis[N], used[N], a[N], b[N];
int maps[N][N]; int found(int u)
{
for(int i = ; i <= n; i++)
{
if(!vis[i] && maps[u][i])
{
vis[i] = ;
if(!used[i] || found(used[i]))
{
used[i] = u;
return true;
}
}
}
return false;
}
int main()
{
int w; while(scanf("%d", &n) != EOF)
{
memset(vis, , sizeof(vis));
memset(used, , sizeof(used));
memset(maps, , sizeof(used));
memset(a, , sizeof(a));
memset(b, , sizeof(b)); int ans = ; for(int i = ; i <= n; i++)
for(int j = ; j <= n; j++)
scanf("%d", &maps[i][j]); for(int i = ; i <= n; i++)
{
memset(vis, , sizeof(vis));
if(found(i))
ans++;
}
if(ans < n)
{
printf("-1\n");
continue;
} w = ;
for(int i = ; i <= n; i++)
{
while(used[i] != i)
{
a[w] = i;
b[w] = used[i];
swap(used[a[w]], used[b[w]]); // 如果该行匹配不是自身，就交换匹配。
w++;
}
}
printf("%d\n", w);
for(int i = ; i < w; i++)
printf("C %d %d\n", a[i], b[i]); }
return ;
}