题目链接:点击打开链接
题目大意:给你一对堆,求是否存在一个四维序列(A,B,C,D)其中A<C,B<D,且A!=C或者B!=D,使A到C的哈夫曼距离等于B到D的哈夫曼距离
解题思路:一开始很头大,各种不会怎么想都超时,二分又想不出,最后才发现m的重要性,m表示范围,也就是有2*m种哈夫曼距离,根据抽屉原理,只要超过2*m必有重复,因此剩下来的情况n最大也就1000,暴搜轻松过。
代码:
#include<iostream>
#include<vector>
#include<cmath>
#include<algorithm>
#include<ctime>
#include "cstdio"
#include "string"
#include "string.h"
#include "map"
#include "bitset"
using namespace std;
struct Node
{
int x, y;
bool friend operator<(Node a, Node b)
{
if (a.x == b.y)
return a.y < b.y;
return a.y < b.y;
}
};
map<Node, int>mp;
vector<Node>node;
int vis[2 * 100000 + 1];
int main()
{
int T,n,m;
scanf("%d", &T);
while (T--)
{
mp.clear();
node.clear();
memset(vis, 0, sizeof(vis));
scanf("%d %d", &n, &m);
long long num = 0;
for (int i = 0;i < n;i++)
{
Node temp;
scanf("%d %d", &temp.x, &temp.y);
if (mp[temp])
continue;
mp[temp] = 1;
node.push_back(temp);
}
num = node.size();
if (num*(num - 1) / 2 > 2 * m)
{
puts("YES");
continue;
}
sort(node.begin(), node.end());
bool flag = false;
for (int i = 0;i < num;i++)
{
for (int j = i + 1;j < num;j++)
{
int dx = abs(node[i].x - node[j].x);
int dy = abs(node[i].y - node[j].y);
//cout << dx << " " << dy << endl;
if (vis[dx + dy])
{
flag = true;
break;
}
vis[dx + dy] = 1;
}
if (flag)
break;
}
if (flag)
puts("YES");
else
puts("NO");
}
return 0;
}