Swap(hdu2819,最大匹配+记录路径)

http://acm.hdu.edu.cn/showproblem.php?pid=2819
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29327#problem/B
B - Swap

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

解析：

题意： 给出一个矩阵，任意两行可以交换，任意两列可以交换，问如何交换是的矩阵的对角线全为1，如果可以达到目标则输出步骤，否则输出-1.

思路：

转化为行和列的最大匹配。

如果最大匹配数小于n则说明不可达到目标

否则记录路径并输出

276 KB 78 ms C++ 1298 B

#include<string.h>
#include<stdio.h>
#include<algorithm>
#include <iostream>
using namespace std;
const int maxn=100+10;
int n;
int vis[maxn],result[maxn];
int map[maxn][maxn];
struct node
{
int r;
int c;
}nd[maxn];
int find(int a)
{
for(int i=1;i<=n;i++)
{
if(map[a][i]&&!vis[i])
{
vis[i]=1;
if(result[i]==0||find(result[i]))
{
result[i]=a;
return 1;
}
}
}
return 0;
}
int main()
{int i,k,j,t;
while(scanf("%d",&n)!=EOF)
{int u,v;
//memset(map,0,sizeof(map));
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
scanf("%d",&map[i][j]);
}
int ans=0;
memset(result,0,sizeof(result));
for(i=1;i<=n;i++)
{
memset(vis,0,sizeof(vis));
if(find(i))
{ans++;
}
}
if(ans<n)
{
printf("-1\n");
continue;
}
k=0;
int t;
for(i=1;i<=n;i++)
{
if(result[i]!=i)
{
for(j=i+1;j<=n;j++)
{
if(result[j]==i)
{
nd[k].r=i;
nd[k++].c=j;
t=result[i];//注意这里交换的是列
result[i]=result[j];
result[j]=t;
}
}
}
}
printf("%d\n",k);
for(i=0;i<k;i++)
{ printf("C %d %d\n",nd[i].r,nd[i].c);
}
}
return 0;
}