題目連結

題意：給出一個數列(n<=300),Permutation a1, a2, …, an is prettier than permutation b1, b2, …, bn, if and only if there exists an integer k (1 ≤ k ≤ n) where a1 = b1, a2 = b2, …, ak - 1 = bk - 1 and ak < bk all holds。然後給出一張表，告訴你每一次可將哪兩個元素可以交換。輸出最優排列。

code:

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)
#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)
 

#define mes(a,x,s)  memset(a,x,(s)*sizeof a[0])
#define mem(a,x)  memset(a,x,sizeof a)
#define ysk(x)  (1<<(x))
typedef long long ll;
//const int INF = 0x3f3f3f3f ;
const int maxn= 300 ;
int n,a[maxn+3];
bool dp[maxn+3][maxn+3];

void floyd()
{
    for1(i,n)  dp[i][i]=1;

    for(int k=1;k<=n;k++)
    {
        for
 
(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                dp[j][i]=dp[i][j]= dp[i][j]|| dp[i][k]&&dp[k][j];
            }
        }

    }

}


void greedy()
{
    for1(i,n) 
    {
        int k=i;
        for(int j=i+1;j<=n;j++) if(a[j]<a[k]&&dp[i][j])
        {
            k=j;
        }

        if(k!=i) swap(a[i],a[k]);
    }

    for1(i,n)
    {
        printf("%d%c",a[i],i==n?'\n':' ');
    }

}
int main()
{
    scanf("%d",&n);
    for1(i,n)  scanf("%d",&a[i]);
    char ch;
    for1(i,n) for1(j,n) 
    {
        scanf(" %c",&ch);
        dp[i][j]= ch=='1'?1:0;
    }
    floyd();
    greedy();

    return 0;
}