ZOJ Problem Set 3632 Watermelon Full of Water

ZOJ Problem Set - 3632

Watermelon Full of Water

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Time Limit: 3 Seconds      Memory Limit: 65536 KB

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Watermelon is very popular in the hot summer. Students in ZJU-ICPC Team also love watermelon very much and they hope that they can have watermelon to eat every day during the summer vacation. Suppose there are n days and every day they can buy only one watermelon. The price of watermelon may be different in each day. Besides, sometimes the watermelon they choose to buy may be very big, which means if they buy this watermelon, they will need several days to eat it up. The students want to spend the minimum money to buy enough watermelon so that they can eat watermelon every day. Can you help them?

Notice: When they buy a new watermelon, if they still have an old watermelon, they will throw the old one into dustbin. For example, suppose they buy a watermelon on the fisrt day, and it needs 4 days to eat up the watermelon. But if they buy a new watermelon on the second day and it needs 2 days to eat up the new watermelon, then they will throw the old one, and they have to buy a new watermelon on the fourth day since they don't have any watermelon to eat on that day.

Input

The input contains multiple test cases ( no more than 200 test cases ).
In each test case, first there is an integer, n ( 1 <= n <=50000 ) , which is the number of summer days.
Then there is a line containing n positive integers with the i_th integer indicating the price of the watermelon on the i_th day.
Finally there is line containing n positive integers with the i_th integer indicating the number of days students need to eat up the watermelon bought on the i_th day.
All these integers are no more than 100000 and integers are seperated by a space.

Output

For each case, output one line with an integer which is the minimum money they must spend so that they can have watermelon to eat every day.

Sample Input

4
10 20 1 40
3 2 3 1

Sample Output

11

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Author: HUANG, Qiao

            
            
            
                Contest: ZOJ Monthly, July 2012
//动规加线段树优化

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 50005
#define Max 1000000000000000
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
using namespace std;
long long Min[N<<2];
int p[N],last[N];
long long dp[N];

void build(int l,int r,int k)
{
    Min[k]=Max;
    if(l==r)
     return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void up(int &k)
{
    Min[k]=min(Min[k<<1],Min[k<<1|1]);
}
long long  val;
void update(int &index,int l,int r,int k)
{
    if(l==r)//更新每天买的西瓜最远可以持续到的天数
    {
        Min[k]=min(val,Min[k]);
        return ;
    }
    int m=(l+r)>>1;
    if(index<=m) update(index,lson);
    else  update(index,rson);
    up(k);
}
long long query(int &L,int &R,int l,int r,int k)
{
    if(L<=l&&R>=r)
    {
        return Min[k];
    }
    int m=(l+r)>>1;
    long long t1=Max,t2=Max;
    if(L<=m) t1=query(L,R,lson);
    if(R>m)  t2=query(L,R,rson);
    return min(t1,t2);
}
int main()
{
    int n;
    int i,k;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
         scanf("%d",&p[i]);
        for(i=1;i<=n;i++)
         scanf("%d",&last[i]);
        build(1,n,1);
        dp[0]=0;
        for(i=1;i<=n;i++)
        {
            k=last[i]+i-1;
            k=k>n?n:k;val=dp[i-1]+p[i];
            update(k,1,n,1);
            dp[i]=query(i,n,1,n,1);//从当前到结束找到最小的
        }
        printf("%lld\n",dp[n]);
    }
    return 0;
}
//还有一种是用优先队列优化，经测试，线段树优化要快点
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 50005
using namespace std;
long long Min[N<<2];
int p[N],last[N];
long long dp[N];
struct node
{
    long val;
    int last;
    bool operator<(const node& a)const
    {
        return val>a.val;
    }
};
int main()
{
    int n;
    int i;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
         scanf("%d",&p[i]);
        for(i=1;i<=n;i++)
         scanf("%d",&last[i]);
       priority_queue<node> q;
          node temp;
        dp[1]=p[1];
        temp.val=p[1]; temp.last=last[1];
        q.push(temp);
       for(i=2;i<=n;i++)
       {
           temp.val=dp[i-1]+p[i];
           temp.last=last[i]+i-1;
           q.push(temp);
           while(q.top().last<i) q.pop();
           dp[i]=q.top().val;
       }
       printf("%lld\n",dp[n]);
    }
    return 0;
}