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+7;
const int MAXN = 100+10;

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

bool dfs(int u)
{
    for(int i = 1; i<=n; i++)
    if(M[u][i] && !vis[i])
    {
        vis[i] = true;
        if(ylink[i]==-1 || dfs(ylink[i]))
        {
            ylink[i] = u;
            xlink[u] = i;
            return true;
        }
    }
    return false;
}

int hungary()
{
    int ret = 0;
    memset(xlink, -1, sizeof(xlink));
    memset(ylink, -1, sizeof(ylink));
    for(int i = 1; i<=n; i++)
    {
        memset(vis, 0, sizeof(vis));
        if(dfs(i)) ret++;
    }
    return ret;
}

int main()
{
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 1; i<=n; i++)
        for(int j = 1; j<=n; j++)
            scanf("%d", &M[i][j]);

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

        int ans = 0, op[MAXN][2];
        for(int i = 1; i<n; i++)
        for(int j = i+1; j<=n; j++)
        {
            if(xlink[j]==i)
            {
                swap(xlink[j], xlink[i]);
                op[++ans][0] = i; op[ans][1] = j;
                break;
            }
        }

        printf("%d\n", ans);
        for(int i = 1; i<=ans; i++)
            printf("R %d %d\n", op[i][0], op[i][1]);
    }
}