# 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

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

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

const int N = 105 ;
int n,ans,tot;

inline void init(){
memset(match,-1,sizeof(match));ans=0;tot=0;
memset(a,0,sizeof(a));memset(b,0,sizeof(b));
}

inline int dfs(int x){
for(int i=1;i<=n;i++){
check[i]=1;
if(match[i]==-1 || dfs(match[i])){
match[i]=x;
return 1;
}
}
}
return 0;
}

inline void Swap(){
for(int i=1;i<=n;i++){
if(match[i]!=i){
a[++tot]=i;b[tot]=match[i];
for(int j=1;j<=n;j++){
if(match[j]==i){
swap(match[i],match[j]);
break ;
}
}
}
}
}

int main(){
while(scanf("%d",&n)!=EOF){
init();
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
scanf("%d",&map[i][j]);
}
}
for(int i=1;i<=n;i++){
memset(check,0,sizeof(check));
if(dfs(i)) ans++;
}
if(ans!=n){
puts("-1");continue ;
}
Swap();
printf("%d\n",tot);
for(int i=1;i<=tot;i++) printf("R %d %d\n",a[i],b[i]);
}
return 0;
}

posted @ 2018-11-06 18:30  heyuhhh  阅读(237)  评论(0编辑  收藏  举报