CF724B. Batch Sort[枚举]-白红宇

CF724B. Batch Sort[枚举]

阅读量：5080 次

发布时间：2019-06-12

本文共 2479 字，大约阅读时间需要 8 分钟。

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).

Examples

input

2 4 1 3 2 4 1 3 4 2

output

YES

input

4 4 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3

output

NO

input

3 6 2 1 3 4 5 6 1 2 4 3 5 6 1 2 3 4 6 5

output

YES

貌似交换行后再交换两个列不会得到更优解 爆枚交换哪两列然后每行贪心

#include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            using namespace std;const int N=25,INF=1e9+5;inline int read(){    char c=getchar();int x=0,f=1;    while(c<'0'||c>'9'){
      if(c=='-')f=-1;c=getchar();}    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}    return x*f;}int n,m,a[N][N];bool check(){    for(int i=1;i<=n;i++){        int cnt=0;        for(int j=1;j<=m;j++) if(a[i][j]!=j) cnt++;        if(cnt>2) return false;    }    return true;}int main(){    n=read();m=read();    for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++) a[i][j]=read();        if(check()) {puts("YES");return 0;}        for(int i=1;i<=m;i++)        for(int j=i+1;j<=m;j++){            for(int z=1;z<=n;z++) swap(a[z][i],a[z][j]);            if(check()) {puts("YES");return 0;}            for(int z=1;z<=n;z++) swap(a[z][i],a[z][j]);            }    puts("NO");}

转载于:https://www.cnblogs.com/candy99/p/6058850.html

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