POJ 1738 An old Stone Game（石子合并 经典）

Posted 2020-09-29 十年换你一句好久不见

An old Stone Game

Time Limit: 5000MS		Memory Limit: 30000K
Total Submissions: 3672		Accepted: 1035

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 <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll a[50006],ans,n,pos;
void check(int x)
{
    int i,j;
    int mx=a[x]+a[x-1];
    ans+=mx;
    for(i=x;i<pos-1;i++)
        a[i]=a[i+1];
    pos--;
    for(j=x-1;j>0 && a[j-1]<mx;j--)
        a[j]=a[j-1];
    a[j]=mx;
    while(j>=2 && a[j-2]<=a[j])
    {
        int y=pos-j;//j后面有多少个数，往前移;
        check(j-1);
        j=pos-y;
    }
}
int main()
{
    while(scanf("%lld",&n)&& n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
        }
        pos=1;ans=0;
        for(int i=1;i<n;i++)
        {
            a[pos++]=a[i];
            while(pos>=3 && a[pos-3]<=a[pos-1])
            {
                check(pos-2);
            }
        }
        while(pos>1) check(pos-1);
        printf("%lld\n",ans);
    }
    return 0;
}