POJ 1651 Multiplication Puzzle 区间DP

 1 //#pragma comment(linker, "/STACK:167772160")//手动扩栈~~~~hdu 用c++交
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <iostream>
 6 #include <queue>
 7 #include <stack>
 8 #include <cmath>
 9 #include <set>
10 #include <algorithm>
11 #include <vector>
12 // #include<malloc.h>
13 using namespace std;
14 #define clc(a,b) memset(a,b,sizeof(a))
15 #define LL long long
16 const int inf = 0x3f3f3f3f;
17 const double eps = 1e-5;
18 const double pi = acos(-1);
19 const LL mod = 1e9+7;
20 const int N = 710;
21 // inline int r(){
22 //     int x=0,f=1;char ch=getchar();
23 //     while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘) f=-1;ch=getchar();}
24 //     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
25 //     return x*f;
26 // }
27 
28 int a[110],dp[110][110];//[i-1,j)的最小分数(i-1可以取，j取不到)
29 int main(){
30     int n;
31     while(~scanf("%d",&n)){
32         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
33             clc(dp,0);
34          for(int len=2;len<n;len++){//枚举长度
35             for(int i=2;i+len<=n+1;i++){
36                 int j=i+len-1;
37                 dp[i][j]=inf;
38                 for(int k=i;k<j;k++){//枚举区间[i-1,j）中最后删除的值
39                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+a[i-1]*a[k]*a[j]);
40                 }
41             }
42         }
43         printf("%d\n",dp[2][n]);
44     }
45     return 0;
46 }