hdu 2819 Swap

Swap

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

题意：交换任意一行或者一列使对角线上都是1。

先判断最大匹配是不是有n个，在交换匹配后的行和列再记录。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#define maxn 110
using namespace std;
typedef struct
{
    int a,b;
} node;
node t[maxn*maxn];
int g[maxn][maxn];
bool used[maxn];
int linker[maxn];
int n;
bool dfs(int u)
{
    for(int i=1; i<=n; i++)
    {
        if(g[u][i]&&!used[i])
        {
            used[i]=true;
            if(linker[i]==-1||dfs(linker[i]))
            {
                linker[i]=u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int res=0;
    memset(linker,-1,sizeof(linker));
    for(int i=1; i<=n; i++)
    {
        memset(used,false,sizeof(used));
        if(dfs(i))res++;
    }
    return res;
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(g,0,sizeof(g));
         memset(t,0,sizeof(t));
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
                scanf("%d",&g[i][j]);
        }
        if(hungary()<n)
        {
            cout<<-1<<endl;
            continue;
        }
        int l=0;
        for(int i=1; i<=n; i++)
        {
            if(linker[i]!=i)
            {
                int j;
                for(j=linker[i]; linker[j]!=i; j=linker[j]);
                t[l].a=i;
                t[l].b=j;
                l++;
                linker[j]=linker[i];
                linker[i]=i;
            }
        }
        cout<<l<<endl;
        for(int i=0;i<l;i++)
            printf("C %d %d
",t[i].a,t[i].b);
    }
}