POJ 3638 Moogle

dp[i][j]表示在i位置放第j个的最小error，简单DP即可

#include <stdio.h>#include <string.h>#include <math.h>#define inf 1e9;int h,in[210];double e[210][210];double dp[210][210];double min(double a,double b){return a>b?b:a;}int main(){int t,T,h,c,i,j,k;scanf("%d",&T);for(t=1;t<=T;t++){scanf("%d %d",&h,&c);for(i=1;i<=h;i++)scanf("%d",&in[i]);for(i=1;i<=h;i++)for(j=i;j<=h;j++){if(j==i){e[i][j]=0;continue;}double tem =(double)(in[j]-in[i])/(j-i);double sum=0;for(int k=i+1;k<j;k++)sum+=fabs(tem*(k-i)+in[i]-in[k]);e[i][j]=sum;}for(i=1;i<=h;i++)for(j=1;j<=c;j++){dp[i][j] = inf;}for(i=2;i<=h;i++){for(j=2;j<=c && j<=i;j++){if(j==2){dp[i][j]=e[1][i];  //第一个位置必须有continue;}for(k=2;k<i;k++){if(k>=j-1){dp[i][j]=min(dp[i][j],dp[k][j-1]+e[k][i]);}}}}printf("%.4f\n",dp[h][c]/h);}}