You are given a square matrix consisting of n rows and n columns. We assume that the rows are numbered from 1 to n from top to bottom and the columns are numbered from 1 to n from left to right. Some cells (n - 1 cells in total) of the the matrix are filled with ones, the remaining cells are filled with zeros. We can apply the following operations to the matrix:
- Swap i-th and j-th rows of the matrix;
- Swap i-th and j-th columns of the matrix.
You are asked to transform the matrix into a special form using these operations. In that special form all the ones must be in the cells that lie below the main diagonal. Cell of the matrix, which is located on the intersection of the i-th row and of the j-th column, lies below the main diagonal if i > j.
Input
The first line contains an integer n (2 ≤ n ≤ 1000) — the number of rows and columns. Then follow n - 1 lines that contain one's positions, one per line. Each position is described by two integers xk, yk (1 ≤ xk, yk ≤ n), separated by a space. A pair (xk, yk) means that the cell, which is located on the intersection of the xk-th row and of the yk-th column, contains one.
It is guaranteed that all positions are distinct.
Output
Print the description of your actions. These actions should transform the matrix to the described special form.
In the first line you should print a non-negative integer m (m ≤ 105) — the number of actions. In each of the next m lines print three space-separated integers t, i, j (1 ≤ t ≤ 2, 1 ≤ i, j ≤ n, i ≠ j), where t = 1 if you want to swap rows, t = 2 if you want to swap columns, and i and j denote the numbers of rows or columns respectively.
Please note, that you do not need to minimize the number of operations, but their number should not exceed 105. If there are several solutions, you may print any of them.
Examples
Input
2 1 2
Output
2 2 1 2 1 1 2
Input
3 3 1 1 3
Output
3 2 2 3 1 1 3 1 1 2
Input
3 2 1 3 2
Output
0
题解： 第一次做的时候傻逼了 看成求最小操作数了 原来前面一个not 既然步数 无所谓 那么我们就一点点模拟进行就好了
因为1 就n-1个 所以一定存在 一行 一列 没有1 那么我们先把没有1的列换到 第n列 然后 把 有1的行换到第n行 我们把1 不断的向左 下 移动 从右往左 不断把有1的列 向左移动 从上往下 不断把有1的行往下移动 当我们先移动行时 就确保了 第n列都为0 所以在往下换行的时候 就确保1 都在第n行的规定返回内 然后n不断减小进行即可 注意的是 当进行完一次n之后 在第n行的1 在后面的操作中是肯定不会换出去的 因为n减小了 所以把这些1 给标记为0 详细看代码 更好懂一点
#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int N=1e3+10;
int n,mp[N][N];
struct node{
int op,x,y;
}a[N*100];
int len;
int r[N],c[N];
int main()
{
int x,y;
while(~scanf("%d",&n))
{
memset(mp,0,sizeof(mp));
memset(r,0,sizeof(r));
memset(c,0,sizeof(c));
len=0;
for(int i=1;i<n;i++)
{
scanf("%d%d",&x,&y);
mp[x][y]++;
r[x]++,c[y]++;
}
for(;n>0;n--)
{
for(int i=1;i<n;i++)
{
if(c[i]==0)
{
for(int j=1;j<=n;j++)
swap(mp[j][i],mp[j][n]);
swap(c[i],c[n]);
a[len].x=i,a[len].y=n,a[len++].op=2;
break;
}
}
for(int i=1;i<n;i++)
{
if(r[i]!=0)
{
for(int j=1;j<=n;j++)
swap(mp[i][j],mp[n][j]);
swap(r[i],r[n]);
a[len].x=i,a[len].y=n,a[len++].op=1;
break;
}
}
for(int i=1;i<=n;i++)
if(mp[n][i])
c[i]--;
}
printf("%d\n",len);
for(int i=0;i<len;i++)
printf("%d %d %d\n",a[i].op,a[i].x,a[i].y);
}
return 0;
}