Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) B. Batch Sort 暴力

B. Batch Sort

题目连接：

http://codeforces.com/contest/724/problem/B

Description

output

standard output

You are given a table consisting of n rows and m columns.

Numbers in each row form a permutation of integers from 1 to m.

You are allowed to pick two elements in one row and swap them, but no more than once for each row. Also, no more than once you are allowed to pick two columns and swap them. Thus, you are allowed to perform from 0 to n + 1 actions in total. Operations can be performed in any order.

You have to check whether it's possible to obtain the identity permutation 1, 2, ..., m in each row. In other words, check if one can perform some of the operation following the given rules and make each row sorted in increasing order.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 20) — the number of rows and the number of columns in the given table.

Each of next n lines contains m integers — elements of the table. It's guaranteed that numbers in each line form a permutation of integers from 1 to m.

Output

If there is a way to obtain the identity permutation in each row by following the given rules, print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).

Sample Input

2 4

1 3 2 4

1 3 4 2

Sample Output

YES

Hint

题意

给你n行，每行都是一个1-m的排列。

你可以交换任意两列，并且你可以每行最多交换两个元素，问你能不能使得每行都是单增的

题解：

暴力枚举嘛，数据范围这么小

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 55;

int a[maxn][maxn],n,m;
bool check(int x,int y)
{
    for(int i=1;i<=n;i++)
    {
        int flag=0;
        for(int j=1;j<=m;j++)
        {
            int tmp=a[i][j];
            if(j==x)tmp=a[i][y];
            if(j==y)tmp=a[i][x];
            if(tmp!=j)flag++;
        }
        if(flag!=0&&flag!=2)return false;
    }
    return true;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
    for(int i=1;i<=m;i++)
        for(int j=1;j<=i;j++)
            if(check(i,j))
                return puts("YES"),0;
    printf("NO");
}