poj 1651 Multiplication Puzzle (区间dp)
题目链接:http://poj.org/problem?id=1651
Description
The goal is to take cards in such order as to minimize the total number of scored points.
For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10*1*50 + 50*20*5 + 10*50*5 = 500+5000+2500 = 8000
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
1*50*20 + 1*20*5 + 10*1*5 = 1000+100+50 = 1150.
Input
Output
Sample Input
6
10 1 50 50 20 5
Sample Output
3650
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define ll long long
const int maxn=1e2+;
const int INF=0x3f3f3f3f; int dp[][];
int a[]; int main()
{
int n;
while(scanf("%d",&n)==)
{
memset(dp,,sizeof(dp));
for(int i=; i<=n; i++) scanf("%d",&a[i]);
for(int d=; d<n; d++)
for(int i=; i+d<=n; i++){
int j=i+d;
dp[i][j]=INF;
for(int k=i+; k<j; k++)
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[k]*a[i]*a[j]);
}
printf("%d\n",dp[][n]);
}
return ;
}
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