HDU 2819 — Swap 二分匹配

Swap

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2174 Accepted Submission(s): 774
Special Judge

[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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”.

[align=left]Sample Input[/align]

2
0 1
1 0
2
1 0
1 0

[align=left]Sample Output[/align]

1
R 1 2
-1

[align=left]Source[/align]
2009 Multi-University Training Contest 1 - Host by TJU
题意：给定一个n*n的01矩阵；通过行或列的变换使得主对角线上都为1；
题解：
　1.第i行放到第j行可以使得第j行的主对角线为1;

2.第j列放到第i列可以使得第j列的主对角线为1;

那么将行作为X集合，列作为Y集合，如果map[i][j]==1，那么Xi->Yj连边，求最大匹配，这样的话没有任何一个行被两个列匹配，

倘若最大匹配为n，即满足题意；

///1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <queue>
#include <typeinfo>
#include <map>
typedef long long ll;
using namespace std;
#define inf 10000000
inline ll read()
{
ll x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
//***************************************************************
int lk[101],vis[101],n;
char mp[101][101];
bool dfs(int x){
for(int i=1;i<=n;i++){
if(mp[x][i]=='1'&&!vis[i])
{
vis[i]=1;
if(lk[i]==0||dfs(lk[i]))
{
lk[i]=x;
return 1;
}
}
}
return 0;

}
int main()
{

while(scanf("%d",&n)!=EOF)
{int x;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
scanf("%d",&x);
mp[i][j]=x+'0';
}
}
memset(lk,0,sizeof(lk));
int ans=0;
for(int i=1;i<=n;i++){
memset(vis,0,sizeof(vis));
if(dfs(i))ans++;
}
int flag;
if(ans<n)printf("-1\n");
else {
cout<<n<<endl;
//for(int i=1;i<=n;i++) cout<<i<<" "<<lk[i]<<endl;
// for(int i=1;i<=n;i++)flk[lk[i]]=i;
for(int i=1;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
if(lk[j]==i){
flag=j;break;
}
}
lk[flag]=lk[i];
cout<<"C "<<i<<" "<<flag<<endl;
}
}
}
return 0;
}

代码