D - Pick The Sticks

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

Description

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

HINT

题意

给你一个长度为l的盒子,然后里面可以放置一排金砖,只要这个金砖的重心在盒子里面就好了

然后问你放置的金砖的最大价值是多少

题解:

不能伸出去,这就是一个01背包

然后我们贪心一下,我们肯定是选择两个金砖砍一半扔边上,然后剩下就是简单的01背包了

于是我们就dp[i][j]表示空间为i,有j个砍了一半扔在了边上的最大价值是多少

然后暴力转移就好了

代码:

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std; long long dp[][][];
struct node
{
int x;
long long y;
};
node p[];
int main()
{
int t;scanf("%d",&t);
for(int cas = ; cas <= t; cas++)
{
memset(dp,,sizeof(dp));
memset(p,,sizeof(p));
int n,l;scanf("%d%d",&n,&l);
long long ans = ;
for(int i=;i<=n;i++)
{
scanf("%d%lld",&p[i].x,&p[i].y);
p[i].x*=;
ans = max(ans,p[i].y);
}
l *= ;
int now = ;
for(int i=;i<=n;i++)
{
now = -now;
for(int j=;j<=l;j++)
for(int k=;k<;k++)
dp[now][j][k]=dp[-now][j][k];
for(int j=l;j>=p[i].x/;j--)
{
for(int k=;k<;k++)
{
if(j>=p[i].x)dp[now][j][k]=max(dp[now][j][k],dp[-now][j-p[i].x][k]+p[i].y);
if(k)dp[now][j][k]=max(dp[now][j][k],dp[-now][j-p[i].x/][k-]+p[i].y);
}
}
}
for(int t=;t<;t++)
for(int i=;i<=l;i++)
for(int k=;k<;k++)
ans = max(dp[t][i][k],ans);
printf("Case #%d: %lld\n",cas,ans);
}
}

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