Swap(二分图的最大匹配)
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?
InputThere 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.OutputFor 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：交换图的某些行或者是某些列，使得这个N*N的图对角线上全部都是1。
题解：首先我们需要明白一个道理,如果通过交换某些行没有办法的到解的话，那么只交换列或者既交换行又交换列那也没办法得到解。为了解这道题，我们需要构造二分图，第一部分X表示的是横坐标，第二部分Y表示纵坐标，如果mp[i][j]==1.那我们就从X的i向Y的j引一条边，那么这条边的含义就可以解释为可以将Y的第j列（因为Y表示的是列的集合）移到第i列，使得a[i][i]变成1，这样就相当于是第i行第i列就变成了1，也就是说对角线多了一个1。这个地方一定要注意理解mp[i][j]的含义！！
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
struct node{
int st;
int ed;
}e[maxn];
int used[maxn];
int a[maxn],b[maxn];
int match[maxn],mp[maxn][maxn];
int vis[maxn][maxn];
int x;
int n,cnt;
int dfs(int u)
{
int i;
for(i=1;i<=n;i++)
{
if(mp[u][i]&&!used[i])
{
used[i]=1;
if(match[i]==-1||dfs(match[i]))
{
match[i]=u;
return 1;
}
}
}
return 0;
}
int hungary()
{
int ans=0,i;
memset(match,-1,sizeof(match));
for(i=1;i<=n;i++)
{
memset(used,0,sizeof(used));
if(dfs(i))
ans++;
}
return ans;
}
int main()
{
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
memset(mp,0,sizeof(mp));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&x);
if(x==1)
{
mp[i][j]=1;
}
}
}
int ans=hungary();
if(ans<n)
printf("-1\n");
else
{
int cnt = 0, a[maxn]={0}, b[maxn]={0};
for(int i=1; i<=n; i++)
{
while(i != match[i])
{
a[cnt] = i;
b[cnt] = match[i];
swap(match[i], match[match[i]]);
cnt ++;
}
}
printf("%d\n", cnt);
for(int i=0; i<cnt; i++)
printf("C %d %d\n", a[i], b[i]);
}
}
}
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
struct node{
int st;
int ed;
}e[maxn];
int used[maxn];
int a[maxn],b[maxn];
int match[maxn],mp[maxn][maxn];
int vis[maxn][maxn];
int x;
int n,cnt;
int dfs(int u)
{
int i;
for(i=1;i<=n;i++)
{
if(mp[u][i]&&!used[i])
{
used[i]=1;
if(match[i]==-1||dfs(match[i]))
{
match[i]=u;
return 1;
}
}
}
return 0;
}
int hungary()
{
int ans=0,i;
memset(match,-1,sizeof(match));
for(i=1;i<=n;i++)
{
memset(used,0,sizeof(used));
if(dfs(i))
ans++;
}
return ans;
}
int main()
{
while(~scanf("%d",&n))
{
memset(vis,0,sizeof(vis));
memset(mp,0,sizeof(mp));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&x);
if(x==1)
{
mp[j][i]=1;
}
}
}
int ans=hungary();
if(ans<n)
printf("-1\n");
else
{
int cnt=0;
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=1;i<=n;i++)
{
while(i!=match[i])
{
a[cnt]=i;
b[cnt]=match[i];
swap(match[i],match[match[i]]);
cnt++;
}
}
printf("%d\n",cnt);
for(int i=0;i<cnt;i++)
{
printf("R %d %d\n",a[i],b[i]);
}
}
}
}
