sgu 183. Painting the balls 动态规划 难度:3

183. Painting the balls

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

Petya puts the N white balls in a line and now he wants to paint some of them in black, so that at least two black balls could be found among any M successive balls. Petya knows that he needs Ci milliliters of dye exactly to paint the i-th ball. Your task is to find out for Petya the minimum amount of dye he will need to paint the balls.

Input

The first line contains two integer numbers N and M (2<=N<=10000, 2<=M<=100, M<=N). The second line contains N integer numbers C1, C2, ..., CN (1<=Ci<=10000).

Output

Output only one integer number - the minimum amount of dye Petya will need (in milliliters).

Sample test(s)

Input

 

 

6 3 
1 5 6 2 1 3

 

 

Output

 

 

9

 

 

Note

Example note: 1, 2, 4, 5 balls must be painted.

思路:dp[i][j]//在i染色,在i-j染色的最小花费 设a b更新到,a b c,由远(距a m-1距c 1)到近以b为中心更新dp即可

 

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int maxn=10001;

const int maxm=101;

const int inf=1e9+5;

int dp[maxn][maxm];//back maxm

int c[maxn];

int n,m;

int main(){

    while(scanf("%d%d",&n,&m)==2){

        //memset(dp,0,sizeof(dp));

        for(int i=0;i<n;i++)scanf("%d",c+i);

        for(int i=0;i<m;i++){

            for(int j=0;j<i;j++){

                dp[i][i-j]=c[i]+c[j];//g[i-j]=min(dp[i][j],g[i-j]);

            }

        }

        for(int j=1;j<n;j++){

            int minn=inf;

            for(int i=min(n,j+m)-1;i>j&&i>=m;i--){

                minn=min(dp[j][m+j-i],minn);

                dp[i][i-j]=minn+c[i];

            }

        }

        int ans=inf;

        for(int i=n-m+1;i<n;i++){

                for(int j=min(m-1,i-n+m);j>0;j--)

            ans=min(ans,dp[i][j]);

        }

        printf("%d\n",ans);

    }

    return 0;

}