SCU3037 Painting the Balls

Description

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 (C_i) 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 (C_1, C_2, ..., C_N (1 le C_i le10000)).

Output

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

Sample Input

6 3
1 5 6 2 1 3

Sample Output

9

(f[i][j])表示最后一个黑球在(i)倒数第二个黑球在(i-j)的最小值。
转移方程$$f[i][j] = min(f[j][1 sim m-j])+C_i$$
前缀优化，另(g[i][j] = min(f[i][1 sim j]))。复杂度(O(NM))。

#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;

const int maxn = 10010,maxm = 110,inf = 1<<30;
int N,M,C[maxn],f[maxn][maxm],g[maxn][maxm],ans = inf;

inline int gi()
{
    char ch; int ret = 0,f = 1;
    do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
    if (ch == '-') f = -1,ch = getchar();
    do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
    return ret*f;
}

int main()
{
	freopen("3037.in","r",stdin);
	freopen("3037.out","w",stdout);
    N = gi(); M = gi();
    for (int i = 1;i <= N;++i) C[i] = gi();
    for (int i = 0;i <= N;++i) for (int j = 0;j <= M;++j) g[i][j] = f[i][j] = inf;
    for (int i = 1;i <= M;++i) for (int j = 1;j < i;++j) f[i][j] = C[i]+C[i-j],g[i][j] = min(g[i][j-1],f[i][j]);
    for (int i = M+1;i <= N;++i) for (int j = 1;j < M;++j) f[i][j] = g[i-j][M-j]+C[i],g[i][j] = min(g[i][j-1],f[i][j]);
    for (int i = 1;i <= M;++i) for (int j = 1;j < i;++j) ans = min(ans,f[N-M+i][j]);
    cout << ans << endl;
    fclose(stdin); fclose(stdout);
    return 0;
}