URAL 1167. Bicolored Horses dp

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1167. Bicolored Horses

Time limit: 1.0 second
Memory limit: 64 MB

Every day, farmer Ion (this is a Romanian name) takes out all his horses, so they may run and play. When they are done, farmer Ion has to take all the horses back to the stables. In order to do this, he places them in a straight line and they follow him to the stables. Because they are very tired, farmer Ion decides that he doesn't want to make the horses move more than they should. So he develops this algorithm: he places the 1st P1 horses in the first stable, the next P2 in the 2nd stable and so on. Moreover, he doesn't want any of the K stables he owns to be empty, and no horse must be left outside. Now you should know that farmer Ion only has black or white horses, which don't really get along too well. If there are i black horses and j white horses in one stable, then the coefficient of unhappiness of that stable is i*j. The total coefficient of unhappiness is the sum of the coefficients of unhappiness of every of the K stables.

Determine a way to place the N horses into the K stables, so that the total coefficient of unhappiness is minimized.

Input

On the 1st line there are 2 numbers: N (1 ≤ N ≤ 500) and K (1 ≤ K ≤ N). On the next N lines there are N numbers. The i-th of these lines contains the color of the i-th horse in the sequence: 1 means that the horse is black, 0 means that the horse is white.

Output

You should only output a single number, which is the minimum possible value for the total coefficient of unhappiness.

Sample

inputoutput

6 3110101

2

Notes

Place the first 2 horses in the first stable, the next 3 horses in the 2nd stable and the last horse in the 3rd stable.

Problem Author: Mugurel Ionut Andreica
Problem Source: Romanian Open Contest, December 2001

将n匹马按照顺序放进k个马槽里，每个马槽里面的不愉快值为颜色为0的马和颜色为1的马的乘积，求怎么放才能使得总的不愉快值最小。

首先预处理出每个区间的马的不愉快值,然后求出总的不愉快值。

dp[i][j]代表在前i个马槽，前j匹马最小的不愉快值，状态转移方程：

dp[i][j]=min(dp[i][j],dp[i-1][k-1]+unhappy[k][j]);  当前马槽的前j匹马最小不愉快值=上一个马槽前k匹马最小不愉快值+将k到j匹马放到第i个马槽的不愉快值.

//0.1252 202 KB#include<stdio.h>#include<string.h>#include<algorithm>#define inf 0x3f3f3f3fusing namespace std;int h[507],dp[507][507],unhappy[507][507];int main(){    int n,kk;    while(scanf("%d%d",&n,&kk)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&h[i]);        memset(unhappy,0,sizeof(unhappy));        for(int i=1;i<=n;i++)dp[0][i]=inf;        int b,w;        for(int i=1;i<=n;i++)//预处理每段区间的不愉快值        {            b=0;w=0;            for(int j=i;j<=n;j++)            {                if(h[j])b++;else w++;                unhappy[i][j]=b*w;            }        }        for(int i=1;i<=kk;i++)//k个马槽            for(int j=1;j<=n;j++)//前j匹马            {                dp[i][j]=inf;                for(int k=1;k<=j;k++)//枚举前j匹马                    dp[i][j]=min(dp[i][j],dp[i-1][k-1]+unhappy[k][j]);            }        printf("%d\n",dp[kk][n]);    }    return 0;}