poj 1738 取石子游戏 区间dpgarsiawachs算法

本文介绍了一个关于石子合并游戏的算法实现,目的是在遵循特定规则的情况下,找到将石子合并为单一堆的最小得分。通过输入石子初始分布情况,程序计算并输出最优得分。

http://poj.org/problem?id=1738

Description

There is an old stone game.At the beginning of the game the player picks n(1<=n<=50000) piles of stones in a line. The goal is to merge the stones in one pile observing the following rules: 
At each step of the game,the player can merge two adjoining piles to a new pile.The score is the number of stones in the new pile. 
You are to write a program to determine the minimum of the total score. 

Input

The input contains several test cases. The first line of each test case contains an integer n, denoting the number of piles. The following n integers describe the number of stones in each pile at the beginning of the game. 
The last test case is followed by one zero. 

Output

For each test case output the answer on a single line.You may assume the answer will not exceed 1000000000.

Sample Input

1
100
3
3 4 3
4
1 1 1 1
0

Sample Output

0
17
8
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
const int N=50005;
int stone[N],ans,j;
void add(int k)
{
    int tem=stone[k]+stone[k-1];
    ans+=tem;
    for(int i=k ; i<j-1; i++)
    {
        stone[i]=stone[i+1];
    }
    j--;
    int m;
    for(m=k-1 ; m>0 && stone[m-1]<tem ; m--)
    {
        stone[m]=stone[m-1];
    }
    stone[m]=tem;
    while(m>=2 && stone[m-2]<=stone[m])
    {
        int d=j-m;
        add(m-1);
        m=j-d;
    }
}
int main()
{
    int n;
    while(cin>>n && n)
    {
        for(int i=0; i<n; i++)
            cin>>stone[i];
        j=1;ans=0;
        for(int i=1; i<n; i++)
        {
            stone[j++]=stone[i];
            while(j>=3 && stone[j-3]<=stone[j-1])
                      add(j-2);
        }
        while(j>1) add(j-1);
        cout<<ans<<endl;
    }
    return 0;
}

算法解释: http://fanhq666.blog.163.com/blog/static/81943426201062865551410/
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