2015南阳CCPC D - Pick The Sticks 背包DP.

时间：2022-06-14 19:28:53

D - Pick The Sticks

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

题意：给你一个长度为L的木棍容器，n个长度a[i],价值v[i]的木棍，将木棍放入容器中， 必须满足：木棍的中心在容器范围内
     也就是说小木棍可以放在边缘最多伸出一半，问你最大价值是多少。
题解：选木棍就是01背包，不过多了一个状态就是：当前是否有伸出去的木棍  只有三种情况： 没有，伸出1个，伸出2个
     我们可以设计：dp[L][3]在长度L内 伸出木棍的情况
     但注意滚动数组，及转移的后效。
///

#include<bits/stdc++.h>

using namespace std ;

typedef long long ll;

#define mem(a) memset(a,0,sizeof(a))

#define meminf(a) memset(a,127,sizeof(a))

#define TS printf("111111\n")

#define FOR(i,a,b) for( int i=a;i<=b;i++)

#define FORJ(i,a,b) for(int i=a;i>=b;i--)

#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define inf 100000

inline ll read()

{

    ll x=,f=;

    char ch=getchar();

    while(ch<''||ch>'')

    {

        if(ch=='-')f=-;

        ch=getchar();

    }

    while(ch>=''&&ch<='')

    {

        x=x*+ch-'';

        ch=getchar();

    }

    return x*f;

}

//****************************************

#define maxn 2005

int v,a,n,L;

ll dp[][],ans,f[][];

int main(){

  int T=read();

  int oo=;

  while(T--){

     scanf("%d%d",&n,&L);

     memset(f,-,sizeof(f));

     memset(dp,-,sizeof(dp));

     f[][]=;

     dp[][]=;

     L*=;

     for(int i=;i<=n;i++){

        scanf("%d%d",&a,&v);a*=;

        for(int j=;j<=L;j++){

            f[j][]=dp[j][];

            f[j][]=dp[j][];

            f[j][]=dp[j][];

        }

        for(int j=;j<=L;j++){

        if(j+a<=L&&f[j][]!=-)

          dp[j+a][]=max(dp[j+a][],f[j][]+v);

          if(j+(a/)<=L&&f[j][]!=-)

          dp[j+(a/)][]=max(dp[j+(a/)][],f[j][]+v);

         if(j+a<=L&&f[j][]!=-)

          dp[j+a][]=max(dp[j+a][],f[j][]+v);

          if(j+a<=L&&f[j][]!=-)

          dp[j+a][]=max(dp[j+a][],f[j][]+v);

          if(j+(a/)<=L&&f[j][]!=-)

          dp[j+(a/)][]=max(dp[j+(a/)][],f[j][]+v);

          if(a>=L&&f[][]!=-)

            dp[L][]=max(dp[L][],f[][]+v);

        }

     }

     ans=-;

    // cout<<dp[12][2]<<endl;

     for(int i=;i<=L;i++){

        ans=max(ans,dp[i][]);

          ans=max(ans,dp[i][]);

            ans=max(ans,dp[i][]);

     }

     printf("Case #%d: ",oo++);

     cout<<ans<<endl;

  }

 return ;

}

代码

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