hdoj 2819 Swap (最大匹配+输出路径)

题意：

给你一个每位是0或1的矩阵，问你是否可以经过交换行和交换列使矩阵的对角线都为1，如果可以输出交换的方法。

思路：

行和列的匹配，左边是行，右边是列，然后如果行列交点是1，那么就可以匹配，看是否为完美匹配，然后输出怎么交换的。

输出交换的路径：

可以锁定行，来选择列和它匹配，将选择的列移动到和该行形成对角线上是1的位置，比如和第一行匹配的列，就要移动要第一列，第二行的，就到第二列。其实就是对第i行，找一个第i个数是1的列和它匹配，然后看是否是最大匹配！

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
vector<int> e[maxn];
int n, match[maxn], vis[maxn], ans1[maxn], ans2[maxn];

bool dfs(int x)
{
for(int i = 0; i < e[x].size(); i++)
{
int to = e[x][i];
if(!vis[to])
{
vis[to] = 1;
if(!match[to] || dfs(match[to]))
{
match[to] = x;
return true;
}
}
}
return false;
}

int Hungary()
{
int res = 0;
for(int i = 1; i <= n; i++)
{
memset(vis, 0, sizeof(vis));
res += dfs(i);
}
return res;
}

int main(void)
{
while(cin >> n)
{
memset(match, 0, sizeof(match));
for(int i = 1; i < maxn; i++)
e[i].clear();
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
{
int t;
scanf("%d", &t);
if(t) e[i].push_back(j);
}
if(Hungary() != n) puts("-1");
else
{
int cnt = 0;
for(int i = 1; i <= n; i++)
{
int pos;
for(int j = 1; j <= n; j++)
if(match[j] == i)
{
pos = j;
break;
}
if(pos != i)
{
ans1[cnt] = i;
ans2[cnt++] = pos;
swap(match[i], match[pos]);
}
}
printf("%d\n", cnt);
for(int i = 0; i < cnt; i++)
printf("C %d %d\n", ans1[i], ans2[i]);
}
}
return 0;
}

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