HDU2819：Swap（二分图匹配）

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4932    Accepted Submission(s): 1836
Special Judge

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2819

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

题意：

通过交换行和列，使左上角到右下角的对角线都为1。

 

题解：

左上角到右下角的对角线坐标为(i,i)，横纵坐标相等，这是一个特性。我们要做的就是通过交换，使1的横纵坐标相等。

考虑最终的情况，对角线上的1横纵坐标都相等，这在二分图中，就相当于1-1,2-2...n-n，相当于两边集合都是平行相连的。

通过这里可以想到二分图的最大匹配就为n，如果小于n，无论怎样，都不能做到一一平行相连。

那么可行性就可以通过二分图最大匹配判定。

输出方案还是挺有意思的，最终的状态是平行相连，那么我就考虑dfs过后的match数组，不水平的连成水平就行了。

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

const int N =  ;
int map[N][N],link[N][N],match[N],check[N],r[N],a[N],b[N];
int n,ans,tot;

inline void init(){
    memset(map,,sizeof(map));memset(link,,sizeof(link));
    memset(match,-,sizeof(match));ans=;tot=;
    memset(a,,sizeof(a));memset(b,,sizeof(b));
}

inline int dfs(int x){
    for(int i=;i<=n;i++){
        if(link[x][i] && !check[i]){
            check[i]=;
            if(match[i]==- || dfs(match[i])){
                match[i]=x;
                return ;
            }
        }
    }
    return ;
} 

inline void Swap(){
    for(int i=;i<=n;i++){
        if(match[i]!=i){
            a[++tot]=i;b[tot]=match[i];
            for(int j=;j<=n;j++){
                if(match[j]==i){
                    swap(match[i],match[j]);
                    break ;
                }
            }
        }
    }
}

int main(){
    while(scanf("%d",&n)!=EOF){
        init();
        for(int i=;i<=n;i++){
            for(int j=;j<=n;j++){
                scanf("%d",&map[i][j]);
                if(map[i][j]) link[i][j]=;
            }
        }
        for(int i=;i<=n;i++){
            memset(check,,sizeof(check));
            if(dfs(i)) ans++;
        }
        if(ans!=n){
            puts("-1");continue ;
        }
        Swap();
        printf("%d\n",tot);
        for(int i=;i<=tot;i++) printf("R %d %d\n",a[i],b[i]);
    }
    return ;
}