摘要： /*动态规划，如何加速呢》？？*/// include file#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <cctype>#include <ctime>#include <iostream>#include <sstream>#include <fstream>#include <iomanip>#include <bitset>#include & 阅读全文

摘要： /*动态规划，半天没看明白DP[i][j] 表示前i个，i是j时的最小代价DP[i][j] = min(DP[i-1][k]+abs(j-k)*c+(j-a[i])^2) 时间复杂度是N*H*H,超时了。。。。对j,k进行分析当j>k, DP[i][j] = min(DP[i-1][k]+j*c-k*c+(j-a[i])^2) = j*c+(j-a[i])^2+min(DP[i-1][k]-k*c);当j<=k,DP[i][j] = min(DP[i-1][k]+k*c-j*c+(j-a[i])^2) = -j*c+(j-a[i])^2+min(DP[i-1][k]+k*c);领悟 阅读全文