hdu 4960 Another OCD Patient(动态规划)

题目：

Another OCD Patient

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1320    Accepted Submission(s): 459

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.
   
   
   
    
  
  

 

Author

SYSU

 

Source

2014 Multi-University Training Contest 9

 

题意：给n个数字，可以把连续的数字合并成一个数字，合并连续X个数字需要cost【X】花费，求把这n个数字转化为回文序列所需的最小花费。

思路：由于最后需要转化成回文序列，我们可以从两端开始模拟，如果左边的数小于右边的数，那么左边加上下一个数，如果左边大于右边，那么右边加上下一个数，直到二者相等，那么这部分就作为一个整体，分别记录下左边和右边合并了多少数字。假设最后左右各有len个部分，但这len个部分还可以从两端或中间合并来得到更小的花费，所以我们dp一下，dp【i】为左右对称前i个部分合并的最小代价，那么dp【i】=min(,dp【j】+cost【左边j到i 部分合并的个数】+cost【右边j到i 部分合并的个数】)，这是从两端的最小花费。从中间合并的时候，分别减去两端合并的个数，再加上中间合并的花费就行了。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/



int a[5005];
int cost[5005];
int dp[5005];
int ll[5005];
int rr[5005];
int main()
{ int n;
while(RI(n)!=EOF)
{   if(!n)
break;
     for(int i=1;i<=n;i++)
      RI(a[i]);
     for(int i=1;i<=n;i++)
        RI(cost[i]);
    int len=0;
    for(int i=1,j=n;i<j;i++,j--)
    {
        long long int lsum=a[i];
        long long int rsum=a[j];
        int lnum=1;
        int rnum=1;

        while(lsum!=rsum)
        {
            if(lsum<rsum)
            {
                lsum+=a[++i];
                lnum++;
            }
            else
            {
                rsum+=a[--j];
                rnum++;
            }
        }
        if(lsum==rsum)
        {len++;
         ll[len]=lnum;
         rr[len]=rnum;
        }


    }
    dp[0]=0;
    for(int i=1;i<=len;i++)
    {
        dp[i]=INT_MAX;
        int lt=0,rt=0;
        for(int j=i;j>=1;j--)
        {  lt+=ll[j];
           rt+=rr[j];

            dp[i]=min(dp[i],dp[j-1]+cost[lt]+cost[rt]);}
    }
    int ans=cost[n];
    for(int i=1;i<=len;i++)
    {
        n-=ll[i]+rr[i];
        ans=min(ans,dp[i]+cost[n]);
    }
    printf("%d\n",ans);

}



        return 0;
}