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 (1≤x<y≤n1≤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;elsecout << "NO" << endl;return 0;
}

转载于:https://www.cnblogs.com/YDDDD/p/10930203.html

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

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

题意：

思路:

代码：

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