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;
else
cout << "NO" << endl; return 0;
}

  

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