Swap(二分图的最大匹配)

Posted 2021-01-13 1013star

Swap

 HDU - 2819 

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?

InputThere 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.OutputFor 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
题意：交换图的某些行或者是某些列，使得这个N*N的图对角线上全部都是1。
题解：首先我们需要明白一个道理,如果通过交换某些行没有办法的到解的话，那么只交换列或者既交换行又交换列那也没办法得到解。为了解这道题，我们需要构造二分图，第一部分X表示的是横坐标，第二部分Y表示纵坐标，如果mp[i][j]==1.那我们就从X的i向Y的j引一条边，那么这条边的含义就可以解释为可以将Y的第j列（因为Y表示的是列的集合）移到第i列，使得a[i][i]变成1，这样就相当于是第i行第i列就变成了1，也就是说对角线多了一个1。这个地方一定要注意理解mp[i][j]的含义！！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
struct node{
    int st;
    int ed;
}e[maxn];
int used[maxn];
int a[maxn],b[maxn];
int match[maxn],mp[maxn][maxn];
int vis[maxn][maxn];
int x;
int n,cnt;
int dfs(int u)
{
    int i;

    for(i=1;i<=n;i++)
    {
        if(mp[u][i]&&!used[i])
        {
            used[i]=1;
            if(match[i]==-1||dfs(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int ans=0,i;

    memset(match,-1,sizeof(match));
    for(i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))
            ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&x);
                if(x==1)
                {
                    mp[i][j]=1;
                }
            }
        }
        int ans=hungary();

        if(ans<n)
            printf("-1
");
        else
        {
            int cnt = 0, a[maxn]={0}, b[maxn]={0};
        for(int i=1; i<=n; i++)
        {
            while(i != match[i])
            {
                a[cnt] = i;
                b[cnt] = match[i];
                swap(match[i],  match[match[i]]);
                cnt ++;
            }
        }

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

    }
}

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
struct node{
    int st;
    int ed;
}e[maxn];
int used[maxn];
int a[maxn],b[maxn];
int match[maxn],mp[maxn][maxn];
int vis[maxn][maxn];
int x;
int n,cnt;

int dfs(int u)
{
    int i;

    for(i=1;i<=n;i++)
    {
        if(mp[u][i]&&!used[i])
        {
            used[i]=1;
            if(match[i]==-1||dfs(match[i]))
            {
                match[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int hungary()
{
    int ans=0,i;

    memset(match,-1,sizeof(match));
    for(i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))
            ans++;
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        memset(mp,0,sizeof(mp));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&x);
                if(x==1)
                {
                    mp[j][i]=1;
                }
            }
        }
        int ans=hungary();

        if(ans<n)
            printf("-1
");
        else
        {
            int cnt=0;
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            for(int i=1;i<=n;i++)
            {
                while(i!=match[i])
                {
                    a[cnt]=i;
                    b[cnt]=match[i];
                    swap(match[i],match[match[i]]);
                    cnt++;
                }
            }
            printf("%d
",cnt);
            for(int i=0;i<cnt;i++)
            {
                printf("R %d %d
",a[i],b[i]);
            }
        }

    }
}