序列合并 优先队列
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题目描述
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N2N^2N2个和,求这N2N^2N2个和中最小的N个。
输入输出格式
输入格式:第一行一个正整数N;
第二行N个整数AiA_iAi?, 满足Ai≤Ai+1A_ile A_{i+1}Ai?≤Ai+1?且Ai≤109A_ile 10^9Ai?≤109;
第三行N个整数BiB_iBi?, 满足Bi≤Bi+1B_ile B_{i+1}Bi?≤Bi+1?且Bi≤109B_ile 10^9Bi?≤109.
【数据规模】
对于50%的数据中,满足1<=N<=1000;
对于100%的数据中,满足1<=N<=100000。
输出格式:输出仅一行,包含N个整数,从小到大输出这N个最小的和,相邻数字之间用空格隔开。
输入输出样例
输入样例#1:
复制
3 2 6 6 1 4 8
输出样例#1: 复制
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; }
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