codeforces266C矩阵化为下三角阵

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我是靠谱客的博主刻苦咖啡，最近开发中收集的这篇文章主要介绍codeforces266C矩阵化为下三角阵，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

C. Below the Diagonal

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 jdenote 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

Copy

2
1 2

output

Copy

2
2 1 2
1 1 2

input

Copy

3
3 1
1 3

output

Copy

input

Copy

3
2 1
3 2

output

Copy

有点小贪心，首先从最后面往前开始换起我们把上面不全为0的行换到下面来这就能保证了第i列之前的列再开始往前换的时候第i行往下的1就不用考虑的因为越往前列的空间越大而且我们是从后往前推的第i列往后的都符合了所以直接把这一行的(其实是这一行对应的1的对应的列减减)。

#include <bits/stdc++.h>
using namespace std;
struct ha
{
    int t,x,y;
}a[1000000];
int b[1010][1010],x[1001],y[1001],cnt=0;
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int c,d;
        cin>>c>>d;
        x[c]++,y[d]++;
        b[c][d]=1;
    }
    for(int i=n;i>0;i--)
    {
        for(int j=1;j<i;j++)
        {
            if(y[j]==0)
            {
                for(int k=1;k<=n;k++)
                    swap(b[k][i],b[k][j]);
                swap(y[i],y[j]);
                a[++cnt].t=2,a[cnt].x=i,a[cnt].y=j;
                break;
            }
        }
        for(int j=1;j<i;j++)
        {
            if(x[j]!=0)
            {
                for(int k=1;k<=n;k++)
                    swap(b[i][k],b[j][k]);
                swap(x[i],x[j]);
                a[++cnt].t=1,a[cnt].x=i,a[cnt].y=j;
                break;
            }
        }
        for(int j=1;j<=n;j++)
        {
            if(b[i][j])y[j]--;
        }
    }
    cout<<cnt<<endl;
    for(int i=1;i<=cnt;i++)
        cout<<a[i].t<<' '<<a[i].x<<' '<<a[i].y<<endl;
 
}