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hdu 4960 Another OCD Patient

2024-07-18 10:38:50   223人阅读

Another OCD Patient

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 87    Accepted Submission(s): 24


Problem Description
Xiaoji is an OCD (obsessive-compulsive disorder) patient. This morning, his children played with plasticene. They broke the plasticene into N pieces, and put them in a line. Each piece has a volume Vi. Since Xiaoji is an OCD patient, he can‘t stand with the disorder of the volume of the N pieces of plasticene. Now he wants to merge some successive pieces so that the volume in line is symmetrical! For example, (10, 20, 20, 10), (4,1,4) and (2) are symmetrical but (3,1,2), (3, 1, 1) and (1, 2, 1, 2) are not.

However, because Xiaoji‘s OCD is more and more serious, now he has a strange opinion that merging i successive pieces into one will cost ai. And he wants to achieve his goal with minimum cost. Can you help him?

By the way, if one piece is merged by Xiaoji, he would not use it to merge again. Don‘t ask why. You should know Xiaoji has an OCD.
 

Input
The input contains multiple test cases.

The first line of each case is an integer N (0 < N <= 5000), indicating the number of pieces in a line. The second line contains N integers Vi, volume of each piece (0 < Vi <=10^9). The third line contains N integers ai (0 < ai <=10000), and a1 is always 0. 

The input is terminated by N = 0.
 

Output
Output one line containing the minimum cost of all operations Xiaoji needs.
 

Sample Input
5
6 2 8 7 1
0 5 2 10 20
0
 

Sample Output
10


Hint
In the sample, there is two ways to achieve Xiaoji‘s goal.
[6 2 8 7 1] -> [8 8 7 1] -> [8 8 8] will cost 5 + 5 = 10.
[6 2 8 7 1] -> [24] will cost 20.
 

Source
2014 Multi-University Training Contest 9
 

题解及代码：

官方题解：

           

思路和官方是一样的，但是比赛时还是没 写出来，比赛后修修改改就过了，好忧桑啊.....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
typedef __int64  ll;
ll org[5010],l[5010],r[5010];
ll val[5010],d[2510];

struct node
{
    int l,r;
} pos[5010];

map<ll,int>m;
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        m.clear();
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&org[i]);
            if(i==1)
                l[i]=org[i];
            else l[i]=l[i-1]+org[i];
            m[l[i]]=i;
        }

        int k=1,flag=1;
        pos[0].l=pos[0].r=0;
        for(int i=n; i>=1; i--)
        {
            if(i==n)
                r[i]=org[i];
            else r[i]=r[i+1]+org[i];
            if(flag&&m.find(r[i])!=m.end())
            {
                pos[k].l=m[r[i]];
                pos[k].r=n-i+1;
                if(pos[k].l+pos[k].r>=n)
                {
                    flag=0;
                }
                k++;
            }
        }
        k--;

        /*for(int i=0;i<=k;i++)
        {
            printf("l:%d r:%d\n",pos[i].l,pos[i].r);
        }
         */
        int mid;
        if(pos[k].l+pos[k].r==n)
        {
            mid=0;
        }
        else
        {
            k--;
            mid=n-pos[k].r-pos[k].l;
        }
        //printf("k:%d mid:%d\n",k,mid);

        val[0]=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%I64d",&val[i]);
        }

        ll ans=-1;
        d[0]=0;
        for(int i=1; i<=k; i++)
        {
            d[i]=val[pos[i].l]+val[pos[i].r];
            for(int j=1; j<i; j++)
            {
                d[i]=min(d[j]+val[pos[i].l-pos[j].l]+val[pos[i].r-pos[j].r],d[i]);
            }
        }

        for(int i=0; i<=k; i++)
        {
            if(ans==-1||ans>d[i]+val[mid+pos[k].l-pos[i].l+pos[k].r-pos[i].r])
                ans=d[i]+val[mid+pos[k].l-pos[i].l+pos[k].r-pos[i].r];
        }
        printf("%I64d\n",ans);
    }
    return 0;
}








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