題目描述

有兩個長度都是N的序列A和B，在A和B中各取一個數相加可以得到N2N^2N2個和，求這N2N^2N2個和中最小的N個。

輸入輸出格式

第一行一個正整數N；

第二行N個整數AiA_iAi?, 滿足Ai≤Ai+1A_i\le A_{i+1}Ai?≤Ai+1?且Ai≤109A_i\le 10^9Ai?≤109;

第三行N個整數BiB_iBi?, 滿足Bi≤Bi+1B_i\le B_{i+1}Bi?≤Bi+1?且Bi≤109B_i\le 10^9Bi?≤109.

【數據規模】

對於50%的數據中，滿足1<=N<=1000；

對於100%的數據中，滿足1<=N<=100000。

輸出僅一行，包含N個整數，從小到大輸出這N個最小的和，相鄰數字之間用空格隔開。

輸入輸出樣例

3
2 6 6
1 4 8

3 6 7

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 700005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == ‘-‘) f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }


/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int  n;
int a[maxn], b[maxn];
map<pii, int>mp;
struct node {
	int x, y;
	bool operator<(const node&t)const {
		if (a[x] + b[y] < a[t.x] + b[t.y])return false;
		return true;
	}
	node(int X, int Y) { x = X, y = Y; }
};
priority_queue<node>q;

int main() {
	//ios::sync_with_stdio(0);
	cin >> n;
	for (int i = 1; i <= n; i++)rdint(a[i]);
	for (int j = 1; j <= n; j++)rdint(b[j]);
	q.push(node(1, 1));
	for (int i = 1; i <= n; i++) {
		while (mp[make_pair(q.top().x, q.top().y)])q.pop();
		int tmpx = q.top().x, tmpy = q.top().y;
		cout << a[tmpx] + b[tmpy] << ‘ ‘;
		mp[make_pair(tmpx, tmpy)] = 1;
		q.push(node(tmpx + 1, tmpy)); q.push(node(tmpx, tmpy + 1));
	}
	
	return 0;
}

序列合並 優先隊列