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  • Codeforces Round #562 (Div. 2) B. Pairs

    链接:https://codeforces.com/contest/1169/problem/B

    题意:

    Toad Ivan has mm pairs of integers, each integer is between 11 and nn, inclusive. The pairs are (a1,b1),(a2,b2),,(am,bm)(a1,b1),(a2,b2),…,(am,bm).

    He asks you to check if there exist two integers xx and yy (1x<yn1≤x<y≤n) such that in each given pair at least one integer is equal to xx or yy.

    思路:

    单独考虑两个完全不相同的对,例如(1,2)-(3,4), 出现这种对时,x和y只能再这两对中取,所以,用vector记录率先出现的一个队,再找没有出现过的队,如果找不到也无所谓,说明一个队里的已经覆盖了全部。

    再对这最多四个值进行枚举对,挨个查找。

    不过别人思路好像跟我不大一样

    代码:

    #include <bits/stdc++.h>
    
    using namespace std;
    
    typedef long long LL;
    const int MAXN = 3e5 + 10;
    const int MOD = 1e9 + 7;
    pair<int, int> node[MAXN];
    int Dis[MAXN];
    int n, m, k, t;
    int p, q, u, v;
    int x, y, z, w;
    
    bool Serch(int a, int b)
    {
        for (int i = 1;i <= m;i++)
        {
            if (node[i].first != a && node[i].first != b && node[i].second != a && node[i].second != b)
                return false;
        }
        return true;
    }
    
    int main()
    {
        cin >> n >> m;
        vector<int> ser;
        bool flag = true;
        for (int i = 1;i <= m;i++)
        {
            cin >> node[i].first >> node[i].second;
            /*
            Dis[node[i].first]++;
            if ((Dis[node[i].first] == 1&& flag))
            {
                ser.push_back(node[i].first);
                if (ser.size() == 4)
                    flag = false;
            }
            Dis[node[i].second]++;
            if ((Dis[node[i].second] == 1&& flag))
            {
                ser.push_back(node[i].second);
                if (ser.size() == 4)
                    flag = false;
            }
            */
            if (ser.size() < 4)
            {
                bool f = true;
                for (int j = 0; j < ser.size(); j++)
                    if (node[i].first == ser[j])
                        f = false;
                for (int j = 0; j < ser.size(); j++)
                    if (node[i].second == ser[j])
                        f = false;
                if (f)
                    ser.push_back(node[i].first), ser.push_back(node[i].second);
            }
        }
        flag = false;
    //    cout << Serch(2, 4) << endl;
    //    for (auto x:ser)
    //        cout << x << ' ' ;
    //    cout << endl;
        for (int i = 0;i < ser.size();i++)
            for (int j = i+1;j < ser.size();j++)
                if (Serch(ser[i], ser[j]))
                    flag = true;
        if (flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/YDDDD/p/10930203.html
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