Swap——hdu 2819

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

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

 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 = 1; i <= n; i++)
    {
        if(!vis[i] && maps[u][i])
        {
            vis[i] = 1;
            if(!used[i] || found(used[i]))
            {
                used[i] = u;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int w;

    while(scanf("%d", &n) != EOF)
    {
        memset(vis, 0, sizeof(vis));
        memset(used, 0, sizeof(used));
        memset(maps, 0, sizeof(used));
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));

        int ans = 0;

        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                scanf("%d", &maps[i][j]);

        for(int i = 1; i <= n; i++)
        {
            memset(vis, 0, sizeof(vis));
            if(found(i))
                ans++;
        }
        if(ans < n)
        {
            printf("-1
");
            continue;
        }

        w = 0;
        for(int i = 1; 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
", w);
        for(int i = 0; i < w; i++)
            printf("C %d %d
", a[i], b[i]);

    }
    return 0;
}