poj 3186 Treats for the Cows(dp)
发布时间:2020-12-14 01:48:42 所属栏目:Linux 来源:网络整理
导读:Description FJ has purchased N (1 = N = 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.? The treats are interes
Description
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.?
The treats are interesting for many reasons:
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1. Input
Line 1: A single integer,N?
Lines 2..N+1: Line i+1 contains the value of treat v(i) Output
Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input 5 1 3 1 5 2 Sample Output 43 Hint
Explanation of the sample:?
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).? FJ sells the treats (values 1,3,1,5,2) in the following order of indices: 1,2,4,making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
?
题意:给你一组序列?只能从左端点?或者 右端点?选取一个数乘上选取它的天数?得到一个最大?权值和
思路: dp[i][j]?表示第i个物品?选取j个左边的物品?的最大权值
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<string> #include<vector> #include<stack> #include<bitset> #include<cstdlib> #include<cmath> #include<set> #include<list> #include<deque> #include<map> #include<queue> #define ll long long int using namespace std; inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;} int moth[13]={0,31,28,30,31}; int dir[4][2]={1,0,1,-1,-1}; int dirs[8][2]={1,1}; const int inf=0x3f3f3f3f; const ll mod=1e9+7; int a[2007]; int dp[2007][2007]; int main(){ ios::sync_with_stdio(false); int n; while(cin>>n){ memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++){ cin>>a[i]; // sum[i]+=a[i]; } dp[1][0]=a[n]; dp[1][1]=a[1]; for(int i=2;i<=n;i++){ for(int j=n;j>=n-i+1;j--) dp[i][0]+=(a[j]*(n-j+1)); for(int j=1;j<=i;j++){ if(dp[i-1][j-1]+a[j]*i<dp[i-1][j]+a[n-(i-1-j)]*i){ dp[i][j]=dp[i-1][j]+a[n-(i-1-j)]*i; }else{ dp[i][j]=dp[i-1][j-1]+a[j]*i; } } } int ans=-inf; for(int i=0;i<=n;i++) ans=max(dp[n][i],ans); cout<<ans<<endl; } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |