ZOJ2501 POJ1976 A Mini Locomotive

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简单的DP

max_coach表示每节小火车头最多可以拉多少节车卡, passenger[i]保存的是从i往前连续max_coach节车卡的人数, 方便运算.

dp[x][y]表示对于第x节车卡, 用了y个火车头后可以拉的最多人数, 则状态转移方程为

dp[x][y]=max(dp[x-1][y], dp[x-max_coach][y-1]+passenger[x]);

dp[x-1][y]表示x节不拉, dp[x-max_coach][y-1]+passenger[x]表示x节拉

/******************************************************************************* # Author : Neo Fung # Email : neosfung@gmail.com # Last modified: 2012-02-24 19:49 # Filename: ZOJ2501 POJ1976 A Mini Locomotive.cpp # Description :  ******************************************************************************/#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX 50010using namespace std;int dp[MAX][4];int temp[MAX],passenger[MAX];int main(void){#ifdef DEBUG    freopen("../stdin.txt","r",stdin);  freopen("../stdout.txt","w",stdout); #endif  int ncases,n,max_coach;scanf("%d",&ncases);while(ncases--){scanf("%d",&n);for(int i=1;i<=n;++i)scanf("%d",&temp[i]);memset(dp,0,sizeof(dp));memset(passenger,0,sizeof(passenger));scanf("%d",&max_coach);for(int i=1;i<=n;++i)if(i>max_coach)passenger[i]=passenger[i-1]+temp[i]-temp[i-max_coach];elsepassenger[i]=passenger[i-1]+temp[i];int sum=0;for(int i=1;i<=n;++i)for(int j=1;j<=3;++j){if(i<=j*max_coach)dp[i][j]=dp[i-1][j]+temp[i];elsedp[i][j]=max(dp[i-1][j], dp[i-max_coach][j-1]+passenger[i]);}printf("%d\n",dp[n][3]);}  return 0;}