Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9697 | Accepted: 6023 |
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
If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
Input
Output
Sample Input
6 10 1 50 50 20 5
Sample Output
3650
Source
Northeastern Europe 2001, Far-Eastern Subregion
簡單的dp
//Serene
//紫書P277 最優矩陣鏈乘 Multiplication Puzzle
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=100+10;
int n,j,num[maxn],dp[maxn][maxn];
int main() {
scanf("%d",&n);
memset(dp,0x7f,sizeof(dp));
for(int i=1;i<=n;++i) scanf("%d",&num[i]);
for(int i=1;i<=n-1;++i) dp[i][i]=0;
for(int len=2;len<=n-1;++len)
for(int i=1;i<=n-len+1;++i) {
j=i+len-1;
for(int k=i;k<=j;++k) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+num[i]*num[j+1]*num[k+1]);
}
cout<<dp[1][n-1];
return 0;
}