CodeForces——948C Producing Snow(优先队列)

题目:

Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.

Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.

Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.

You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.

Input

The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.

The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.

The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.

Output

Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.

 

首先求出每天融化的前缀和，代表改天总共融化的雪量。建立递增的优先队列（队首最小），通过比较snow[i]+sum[i-1]（第i天要使雪堆融化的雪量）与sum[i]（实际到该天一共融化的雪量），若小于则说明融化完毕弹出优先队列，否则等待第x天使得sum[x]>snow[i]+sum[i-1]（i是以前两者相加时的i）。

#include <iostream>
#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
long long sum[1000];
long long t[1000];
int snow[1000];
int main()
{
    int n;
    cin>>n;
    priority_queue<long long,vector<long long>,greater<long long> > p;
    for(int i=1;i<=n;i++)
        cin>>snow[i];
    for(int i=1;i<=n;i++)
        cin>>t[i];
    for(int i=1;i<=n;i++)
        sum[i]=t[i]+sum[i-1];
        long long ans;
    for(int i=1;i<=n;i++){
        ans=0;
        p.push(snow[i]+sum[i-1]);
        while(!p.empty()&&sum[i]>=p.top()){
            ans+=p.top()-sum[i-1];
            p.pop();
        }
        ans+=p.size()*t[i];
        cout<<ans<<endl;

    }

    return 0;
}