Pick The Sticks HDU - 5543【变形01背包】

The story happened long long ago. One day, Cao Cao made a special order called “Chicken Rib” to his army. No one got his point and all became very panic. However, Cao Cao himself felt very proud of his interesting idea and enjoyed it.

Xiu Yang, one of the cleverest counselors of Cao Cao, understood the command Rather than keep it to himself, he told the point to the whole army. Cao Cao got very angry at his cleverness and would like to punish Xiu Yang. But how can you punish someone because he’s clever? By looking at the chicken rib, he finally got a new idea to punish Xiu Yang.

He told Xiu Yang that as his reward of encrypting the special order, he could take as many gold sticks as possible from his desk. But he could only use one stick as the container.

Formally, we can treat the container stick as an L length segment. And the gold sticks as segments too. There were many gold sticks with different length ai and value vi. Xiu Yang needed to put these gold segments onto the container segment. No gold segment was allowed to be overlapped. Luckily, Xiu Yang came up with a good idea. On the two sides of the container, he could make part of the gold sticks outside the container as long as the center of the gravity of each gold stick was still within the container. This could help him get more valuable gold sticks.

As a result, Xiu Yang took too many gold sticks which made Cao Cao much more angry. Cao Cao killed Xiu Yang before he made himself home. So no one knows how many gold sticks Xiu Yang made it in the container.

Can you help solve the mystery by finding out what’s the maximum value of the gold sticks Xiu Yang could have taken?
Input
The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow. Each test case start with two integers, N(1≤N≤1000) and L(1≤L≤2000), represents the number of gold sticks and the length of the container stick. N lines follow. Each line consist of two integers, ai(1≤ai≤2000) and vi(1≤vi≤109), represents the length and the value of the ith gold stick.
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum value of the gold sticks Xiu Yang could have taken.
Sample Input
4

3 7
4 1
2 1
8 1

3 7
4 2
2 1
8 4

3 5
4 1
2 2
8 9

1 1
10 3
Sample Output
Case #1: 2
Case #2: 6
Case #3: 11
Case #4: 3

大概题意

有一个长为L的条形容器，有n个长为ai价值为vi的金条，问把这些金条放进这个条形容器里可以得到的最大价值是多少，金条不能重叠，边上可以露出来一些（分析一下样例就知道）

解题思路

这个题和普通的01背包不同在于有一些物品是可以露出来的，那么我们就在之前的dp方程的基础上在加一重dp[i][k],这个k代表了又几个边上有金条露出来
首先这个题当金条可以直接放进去的时候我们的dp方程和之前的是一样的;

dp[j][k]=max(dp[j][k],dp[j-e[i].l][k]+e[i].v);

当金条要漏出来一些的时候，状态转移方程为

dp[j][k]=max(dp[j][k],dp[j-e[i].l/2][k-1]+e[i].v);

找出这之中的最大值就是答案。
比赛的时候想到了是01背包来着，但是这个露出来的地方不知道怎么处理，还是太菜了，要好好补一下dp相关的知识了

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    int l;
    long long v;
};
node e[100005];
long long dp[20005][4];
long long F(long long a,long long b)
{
    if(a>b)
        return a;
    return b;
}
int main()
{
    int T,p=1;;
    scanf("%d",&T);
    while(T--)
    {
        int n,L,x;
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&L);
        L*=2;
        long long maxx=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d %lld",&e[i].l,&e[i].v);
            e[i].l*=2;
            maxx=F(maxx,e[i].v);
        }

        for(int i=1; i<=n; i++)
        {
            for(int j=L; j>=e[i].l/2; j--)
            {
                for(int k=0; k<3; k++)
                {
                    if(j>=e[i].l)
                        dp[j][k]=F(dp[j][k],dp[j-e[i].l][k]+e[i].v);//max函数不支持long long，用的自定可以函数
                    if(k)
                        dp[j][k]=F(dp[j][k],dp[j-e[i].l/2][k-1]+e[i].v);
                    maxx=F(maxx,dp[j][k]);
                }
            }
        }
        printf("Case #%d: %lld\n",p++,maxx);
    }
}