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  • URAL(timus) 1280 Topological Sorting(模拟)

    Topological Sorting

    Time limit: 1.0 second
    Memory limit: 64 MB
    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 “si ui” (1 ≤ si, uiN; i = 1, 2, …, M), which means that subject si must be studied before subject ui.
    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 si, ui, which describe the order of subjects: subject si must be studied before ui. 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.

    Samples

    inputoutput
    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
    
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <time.h>
    #include <string>
    #include <map>
    #include <stack>
    #include <vector>
    #include <set>
    #include <queue>
    #define inf 10000000
    #define mod 10000
    typedef long long ll;
    using namespace std;
    const int N=1005;
    const int M=100005;
    int power(int a,int b,int c){int ans=1;while(b){if(b%2==1){ans=(ans*a)%c;b--;}b/=2;a=a*a%c;}return ans;}
    int in[N],vis[N],w[N][N];
    int n,m,k;
    vector<int>vec[N];
    
    int main()
    {
        int a,b;
        bool flag=true;
        scanf("%d%d",&n,&m);
        while(m--){
            scanf("%d%d",&a,&b);
            if(!w[a][b]){
                w[a][b]=1;
                vec[a].push_back(b);
            }
        }
        for(int i=1;i<=n;i++){
            scanf("%d",&in[i]);
        }
        for(int i=1;i<=n;i++){
            int cnt=0;
            for(int j=i+1;j<=n;j++){
                if(w[in[i]][in[j]])cnt++;
            }
        if(cnt!=vec[in[i]].size())flag=false;
    
        }
        if(flag)printf("YES
    ");
        else printf("NO
    ");
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/jianrenfang/p/5856205.html
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