一道区间dp的水题
题目链接
思路
很简单的区间dp,思路和floyed差不多，就是需要把项链处理成环形
用(f[l][r])表示以(a[l])开头(a[r])结尾的数串的最大和
转移方程:
[f[l][r]=max(f[l][r],f[l][k]+f[k][r]+a[l] cdot a[k] cdot a[r])
]
代码
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<string>
#include<cstring>
using namespace std;
inline int read() {
char c = getchar();
int x = 0, f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int ans;
int f[1001][1001],n,a[10010];
signed main() {
cin>>n;
for(int i=1; i<=n; ++i) {
cin>>a[i];
a[n+i]=a[i];/*处理成环*/
}
for(int i=2; i<=n+1; ++i) {
for(int l=1; l+i-1<=2*n/*注意应为2*n,因为上面处理成环了*/; ++l) {
int r=l+i-1;
for(int k=l+1; k<=l+i-2; ++k) {
f[l][r]=max(f[l][r],f[l][k]+f[k][r]+a[l]*a[k]*a[r]);
}
}
}
for(int i=1; i<=n; ++i) {
ans=max(ans,f[i][n+i]);
}
cout<<ans;
return 0;
}