洛谷 P1631 序列合并

题意简述

有两个长度都是N的序列A和B，在A和B中各取一个数相加可以得到N^{2个和，求这N}2个和中最小的N个。

题解思路

大根堆，先存入n个和，再比较大小，改变堆中元素。

代码

#include <cstdio>
#include <algorithm>
using namespace std;
int n, a[100001], b[100001], ans[100001];
int cnt, heap[100001];
void up(int x);
void down(int x);
void push(int x);
void pop();
int main()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	for (int i = 1; i <= n; ++i)
		scanf("%d", &b[i]);
	for (int i = 1; i <= n; ++i)
		push(a[1] + b[i]);
	for (int i = 2; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			if (a[i] + b[j] < heap[1])
			{
				pop();
				push(a[i] + b[j]);
			}
			else break;
	for (int i = 1; i <= n; ++i)
	{
		ans[n - i + 1] = heap[1];
		pop();
	}
	for (int i = 1; i <= n - 1; ++i)
		printf("%d ", ans[i]);
	printf("%d\n", ans[n]);
	return 0;
}
void up(int x)
{
	if (x == 1)	return;
	if (heap[x / 2] < heap[x])
	{
		swap(heap[x / 2], heap[x]);
		up(x / 2);
	}
}
void down(int x)
{
	x *= 2;
	if (x > cnt) return;
	if (x < cnt && heap[x + 1] > heap[x]) x++;
	if (heap[x / 2] < heap[x])
	{
		swap(heap[x / 2], heap[x]);
		down(x);
	}
}
void push(int x)
{
	heap[++cnt] = x;
	up(cnt);
}
void pop()
{
	heap[1] = heap[cnt --];
	down(1);
}