四边形不等式优化DP——石子合并问题 学习笔记

  #include <string.h>
#include <iostream>
#include <list>
#include <map>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <queue>
#define  LL long long
#define MOD 1000000007

using namespace std;

const int INF = 0x3f3f3f;
const int N = 1005;

int m[N][N]; //m[i,j]表示合并d[i..j]所得到的最小得分
int s[N][N];  //s[i,j]记录合并方案
int sum[N]; //前i个石子堆石子数量和
int n;

int solve()
{
    for(int i=1; i<=n; i++)
    {
        m[i][i] = 0;   //边界条件
        s[i][i] = i;
    }
        for(int l=1; l<n; l++)  //枚举 l = j‘ - i
        {
            for(int i=1; i+l<=n; i++)
            {
                int j = i+l;
                int tmp = INF;
                int k = 0;
                for(int div=s[i][j-1]; div<=s[i+1][j];div++) //最优决策的单调性
                {
                    if(tmp > m[i][div] + m[div+1][j] + sum[j] - sum[i-1] )
                    {
                        tmp = m[i][div] + m[div+1][j] + sum[j] - sum[i-1];
                        k = div;
                    }
                }
                m[i][j] = tmp;
                s[i][j] = k;
            }
        }
        return m[1][n];
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        sum[0] = 0;
        for(int i=1; i<=n; i++)
        {
            int a;
            scanf("%d",&a);
            sum[i] = sum[i-1] + a;
        }
        printf("%d\n",solve());
    }
    return 0;
}