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 110using 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\n",t[i].a,t[i].b);    }}