3002 stone merge 3, stone merge

3002 stone merge 3, stone merge
Time Limit: 1 s space limit: 256000 KB title level: DiamondQuestionView running resultsDescriptionDescription

There are n piles of stones in a column, each pile of stones has a weight w [I], each merge can combine adjacent two piles of stones, the cost of one merge is the weight of two piles of stones and w [I] + w [I + 1]. Ask how to arrange the merge sequence to minimize the total merge cost.

Input description Input Description

The first line is an integer n (n <= 3000)

N integers w1, w2. .. wn (wi <= 3000) in the second row)

Output description Output Description

An integer indicates the minimum merge cost.

Sample Input Sample Input

4

4 1 1 4

Sample output Sample Output

18

Data range and prompt Data Size & Hint

The data range is extended compared to the "Stone merge"

CATEGORY tag Tags click here to expandDynamic Programming interval-type DP monotonic DP enumeration Length + quadrilateral inequality Optimization

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #define lli long long int  7 using namespace std; 8 const int MAXN=5001; 9 const int maxn=0x7fffffff;10 void read(int &n)11 {12     char c='+';int x=0;bool flag=0;13     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}14     while(c>='0'&&c<='9')15     x=(x<<1)+(x<<3)+c-48,c=getchar();16     flag==1?n=-x:n=x;17 }18 int n;19 int a[MAXN]; 20 int sum[MAXN];21 int dp[MAXN][MAXN];22 int mid[MAXN][MAXN];23 int  main()24 {25     read(n);26     for(int i=1;i<=n;i++)27         read(a[i]);28     for(int i=1;i<=n;i++)29         sum[i]=a[i]+sum[i-1];30     for(int i=1;i<=n;i++)31         dp[i][i]=0,mid[i][i]=i;32     for(int len=1;len<=n-1;len++)33     {34         for(int i=1;i<=n-len;i++)35         {36             int j=len+i;37             dp[i][j]=maxn;38             for(int k=mid[i][j-1];k<=mid[i+1][j];k++)39             {40                 if(dp[i][j]>dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1])41                 {42                     dp[i][j]=dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1];43                     mid[i][j]=k;44                 }45             }46         }47     }48     49     printf("%d",dp[1][n]);50     return 0;51 }