dp-01背包问题 （升级）

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?

InputThe 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.OutputFor 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

        
 

Hint

In the third case, assume the container is lay on x-axis from 0 to 5. Xiu Yang could put the second gold stick center at 0 and put the third gold stick center at 5, 
so none of them will drop and he can get total 2+9=11 value. 

In the fourth case, Xiu Yang could just put the only gold stick center on any position of [0,1], and he can get the value of 3.

        
题目大意 ：
　　有一个长度为 l 的容器 ，往容器内存放长度不同的东西，只要保证所放的物品其重心在容器上，此物品就可以放在上边，如果是只有一个物品的时候，不管物品有多长，物品都可以放在上边，简单想的话，这题就是一个01背包的问题，在面对一个物品的时候我可以选择放或不放，但是其唯一有些不同的地方，
我可以把此物品的一半放在容器中，那么就会多出来几种情况，即有 0 个物品在两侧 ，有 1 个物品在两侧 ， 有 2 个物品在两侧 。那么显然此时就可以将普通的 01背包一维数组升级成一个二维的 。
　　注意此题还有一个要注意的地方，就是我如果在考虑一个物品是一半在容器中，那么此物品长的一半就有可能为浮点数，所以我可以进行一个小操作 ， 就是将容器的长度 和 所输入的所有物品长扩大一倍 。
　　注意输出用 long long 。

代码示例 ：
　　

/*
 * Author:  renyi 
 * Created Time:  2017/8/29 21:46:26
 * File Name: 
 */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
using namespace std;
const int maxint = -1u>>1;
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define ll long long

int we[1005];
ll dp[5005][5] ,va[1005] ;

int main() {
    int t , n , l ;
    
    cin >> t;
    int f = 1;
    while ( t-- ) {
        scanf ( "%d%d" , &n , &l ) ;
        l *= 2 ;
        ll maxn = 0; 
        for (int i = 1; i <= n; i++) {
            scanf ("%d%lld", &we[i], &va[i]);
            we[i] *= 2;
            maxn = Max(maxn, va[i]);
        }
  //      printf ( "**
" ) ;
        memset ( dp, 0, sizeof(dp));
           
        for (int i = 1; i <= n; i++){
            for (int j = l; j >= we[i]/2; j--){
                for(int k = 0; k < 3; k++) {
                    if (j>=we[i])
                        dp[j][k] = Max(dp[j][k], dp[j-we[i]][k] + va[i]);
                    if (k > 0){
                        dp[j][k] = Max(dp[j][k], dp[j-we[i]/2][k-1] + va[i]);
                    } 
            //        maxn = Max(maxn, dp[j][k]);
                }
            }
        }
 //       int maxn = 0;
        for (int i = l; i >= 0; i--){
            for (int k = 0; k < 3; k++){
                maxn = Max(maxn, dp[i][k]);
            }
        }
        
        printf("Case #%d: %lld
", f++, maxn) ;
    }
    return 0;
}