ACM: 动态规划题 poj&nb…

An old StoneGame

Description

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

Input

The input contains several testcases. The first line of each test case contains an integer n,denoting the number of piles. The following n integers describe thenumber 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 theanswer on a single line.You may assume the answer will not exceed1000000000.

Sample Input

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

Sample Output

0
17
8

题意: 合并石子问题, 将相邻两堆石子合并, 每次得分是合并成新的一堆石子个数, 最后累加最小值.

     nkoj 1137也是一类题, 规模比这个小.

解题思路:

     1. 这类题目一开始想到是DP, 设dp[i][j]表示第i堆石子到第j堆石子合并最小得分.

        状态方程: dp[i][j] = min(dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]);

         sum[i]表示第1到第i堆石子总和.递归记忆化搜索即可.

      2.不过此题有些不一样, 1<=n<=50000范围特大,dp[50000][50000]开不到这么大数组.

        问题分析:

         (1).假设我们只对3堆石子a,b,c进行比较, 先合并哪2堆, 使得得分最小.

             score1 = (a+b) + ( (a+b)+c )

             score2 = (b+c) + ( (b+c)+a )

             再次加上score1 <= score2, 化简得: a <= c,可以得出只要a和c的关系确定,

             合并的顺序也确定.

         (2).GarsiaWachs算法, 就是基于(1)的结论实现.找出序列中满足stone[i-1] <=

             stone[i+1]最小的i, 合并temp = stone[i]+stone[i-1], 接着往前面找是否

             有满足stone[j] > temp, 把temp值插入stone[j]的后面(数组的右边).循环

             这个过程一直到只剩下一堆石子结束.

       (3). 为什么要将temp插入stone[j]的后面,可以理解为(1)的情况

            从stone[j+1]到stone[i-2]看成一个整体 stone[mid],现在stone[j],

            stone[mid], temp(stone[i-1]+stone[i-1]), 情况因为temp <stone[j]

             ,因此不管怎样都是stone[mid]和temp先合并,所以讲temp值插入stone[j]

            的后面是不影响结果.

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 50005

int n;
int a[MAX];
int num, result;

void combine(int k)
{
 int i, j;
 int temp = a[k]+a[k-1];
 result += temp;
 for(i = k; i < num-1; ++i)
  a[i] = a[i+1];
 num--;
 for(j = k-1; j > 0&& a[j-1] < temp;--j)
  a[j] = a[j-1];
 a[j] = temp;
 while(j >= 2&& a[j] >=a[j-2])
 {
  int d = num-j;
  combine(j-1);
  j = num-d;
 }
}

int main()
{
 int i;
// freopen("input.txt", "r", stdin);
 while(scanf("%d", &n) !=EOF)
 {
  if(n == 0) break;
  for(i = 0; i <n; ++i)
   scanf("%d",&a[i]);
  num = 1;
  result = 0;
  for(i = 1; i <n; ++i)
  {
   a[num++] =a[i];
   while(num>= 3 && a[num-3]<= a[num-1])
    combine(num-2);
  }

  while(num >1) combine(num-1);
  printf("%d\n", result);
 }
 return 0;
}