icpc测试赛[E]Eat Walnuts

https://ac.nowcoder.com/acm/contest/8688/E

给定一串数字a b c d e，删除某个数的代价是左右和的平方，eg: 删b，代价是(a+b+c)² ，求最小代价

解： dp[i][j]代表从i~j的最小代价，不会删除i和j（因为靠边凑不了3个）

　　dp[i][j]=min(dp[i][j],dp[i][h]+dp[h][j]+(a[i]+a[h]+a[j])²)

　　注意dp时从区间长度开始递增

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<algorithm>
 4 #include<cstring> 
 5 #include<queue>
 6 #include<iostream>
 7 #include<vector>
 8 #include<iostream>
 9 using namespace std;
10 #define ll long long
11 int const N=120;
12 int const INF=2147483647;
13 ll a[N],f[N][N];
14 int n;
15 int main(){
16   while (scanf("%d",&n)!=EOF) {
17       for (int i=0;i<n;i++)
18         for (int j=0;j<n;j++)
19           f[i][j]=INF;
20       for (int i=0;i<n;i++){
21       scanf("%lld",&a[i]);
22       f[i][i+1]=0;
23       }
24       for (int h=3;h<=n;h++)
25         for (int i=0;i<=n-h;i++){
26             int j=i+h-1;
27             for (int k=i+1;k<j;k++){
28               ll temp=(a[i]+a[j]+a[k])*(a[i]+a[j]+a[k]);
29               f[i][j]=min(f[i][j],temp+f[i][k]+f[k][j]);
30             }
31           }
32     printf("%lld\n",f[0][n-1]);
33   }
34 return 0;
35 }

标签：const,Walnuts,int,long,icpc,include,代价,Eat,dp
来源： https://www.cnblogs.com/ikkvix/p/13917465.html