POJ1976A Mini Locomotive(01背包装+连续线段长度）

A Mini Locomotive

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 2485		Accepted: 1388

Description

A train has a locomotive that pulls the train with its many passenger coaches. If the locomotive breaks down, there is no way to pull the train. Therefore, the office of railroads decided to distribute three mini locomotives to each station. A mini locomotive can pull only a few passenger coaches. If a locomotive breaks down, three mini locomotives cannot pull all passenger coaches. So, the office of railroads made a decision as follows:

1. Set the number of maximum passenger coaches a mini locomotive can pull, and a mini locomotive will not pull over the number. The number is same for all three locomotives.
2. With three mini locomotives, let them transport the maximum number of passengers to destination. The office already knew the number of passengers in each passenger coach, and no passengers are allowed to move between coaches.
3. Each mini locomotive pulls consecutive passenger coaches. Right after the locomotive, passenger coaches have numbers starting from 1.

For example, assume there are 7 passenger coaches, and one mini locomotive can pull a maximum of 2 passenger coaches. The number of passengers in the passenger coaches, in order from 1 to 7, is 35, 40, 50, 10, 30, 45, and 60.

If three mini locomotives pull passenger coaches 1-2, 3-4, and 6-7, they can transport 240 passengers. In this example, three mini locomotives cannot transport more than 240 passengers.

Given the number of passenger coaches, the number of passengers in each passenger coach, and the maximum number of passenger coaches which can be pulled by a mini locomotive, write a program to find the maximum number of passengers which can be transported by the three mini locomotives.

Input

The first line of the input contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows:
The first line of the input file contains the number of passenger coaches, which will not exceed 50,000. The second line contains a list of space separated integers giving the number of passengers in each coach, such that the i^th number of in this line is the number of passengers in coach i. No coach holds more than 100 passengers. The third line contains the maximum number of passenger coaches which can be pulled by a single mini locomotive. This number will not exceed 1/3 of the number of passenger coaches.

Output

There should be one line per test case, containing the maximum number of passengers which can be transported by the three mini locomotives.

Sample Input

1
7
35 40 50 10 30 45 60
2

Sample Output

240
题意：

有三个火车头，n个车厢，每个车厢里面对应的有一定的人数。规定每个火车头最多拉m个连续的车厢而且他们拉的车厢一定是从左到右连续的，问它能够拉的最多的人数；

思路：

类似01背包的解法，首先每个火车最多拉m个连续的车厢，这里我们把只要存在连续的m个车厢的就看成一个物品。相当于往背包容量为3的背包里面放物品所得的最大价值量。但是这里注意每连续的m个车厢为一个物品，f[i][j] = max(f[i - 1][j],f[i - m][j - 1] + sum[i] - sum[i - m]); 这里对于每个物品要么不放，要么就是放（放连续的m个车厢）

sum[i] = a[0] + a[1] + ...  + a[i];

之前看到这个解题思路感觉一点有疑问：会不会有重复，第一节拉1，2；第二节拉2,3这样的，最后结论是不会;因为不取这个车厢的话，那必然就是【i-1】【j】,如果取的话那么肯定就是【i-m】【j-1】,跳到了i-m了，所以不会重，太弱了，其实这道题也挺简单，就是不会，弱

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int MAX = 50000 + 10;
int dp[MAX][5],sum[MAX],a[MAX];
int main()
{
    int t,n,m;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        memset(dp, 0, sizeof(dp));
        memset(sum, 0, sizeof(sum));
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++)
            sum[i] = sum[i - 1] + a[i];
        scanf("%d", &m);
        int tp;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= 3; j++)
            {
                if(i < m)    //因为最多是m节，不足m也是可以的，需要处理一下
                {
                    tp = 0;
                }
                else
                    tp = i - m;
                dp[i][j] = max(dp[i - 1][j], dp[tp][j - 1] + sum[i] - sum[tp]);
            }
        }
        printf("%d
", dp[n][3]);
    }
    return 0;
}