Topological Sorting URAL - 1280_ural哈夫曼树-优快云博客

本文链接：https://blog.youkuaiyun.com/sdut_jk17_zhangming/article/details/80765845

本文介绍了一种验证编程学习计划正确性的方法。针对特定的学科限制条件，利用拓扑排序来判断给定的学习顺序是否合理。适用于编程竞赛训练场景。

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Michael wants to win the world championship in programming and decided to study N subjects (for convenience we will number these subjects from 1 to N). Michael has worked out a study plan for this purpose. But it turned out that certain subjects may be studied only after others. So, Michael’s coach analyzed all subjects and prepared a list of M limitations in the form “ s_i u_i” (1 ≤ s_i, u_i ≤ N; i = 1, 2, …, M) , which means that subject s_i must be studied before subject u_i.

Your task is to verify if the order of subjects being studied is correct.

Remark. It may appear that it’s impossible to find the correct order of subjects within the given limitations. In this case any subject order worked out by Michael is incorrect.

Limitations
1 ≤ N ≤ 1000; 0 ≤ M ≤ 100000.

Input

The first line contains two integers N and M ( N is the number of the subjects, M is the number of the limitations). The next M lines contain pairs s_i, u_i, which describe the order of subjects: subject s_i must be studied before u_i. Further there is a sequence of N unique numbers ranging from 1 to N — the proposed study plan.

Output

Output a single word “YES” or “NO”. “YES” means that the proposed order is correct and has no contradictions with the given limitations. “NO” means that the order is incorrect.

Example

input	output
5 6 1 3 1 4 3 5 5 2 4 2 1 2 1 3 4 5 2	YES
5 6 1 3 1 4 3 5 5 2 4 2 1 2 1 2 4 5 3	NO

题意

给你一些两个数之间的先后顺序，问某个顺序是否成立

思路

拓扑排序

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[20];
ll flag;
ll w;
void f(ll x)
{
    ll i;
    if(x==1)
    {
        flag=1;
        return ;
    }
    for(i=9;i>=2;i--)
    {
        if(x%i==0)
        {
            a[i]++;
            f(x/i);
            return ;
        }
    }
}
int main()
{
    ll n,i;
    flag=0;
    cin>>n;
    w=0;
    if(n==0)
    {
        printf("10\n");
        return 0;
    }
    if(n==1)
    {
        printf("1\n");
        return 0;
    }
    memset(a,0,sizeof(a));
    f(n);
//    for(i=2;i<=9;i++)
//        printf("%d ",a[i]);
    if(!flag) printf("-1\n");
    else
    {
        for(i=2;i<=9;i++)
        {
            while(a[i])
            {
                printf("%lld",i);
                a[i]--;
            }
        }
        printf("\n");
    }
    return 0;
}