2015南阳CCPC Hdu5547

D - Pick The Sticks
Time Limit:10000MS    Memory Limit:65535KB    64bit IO Format:%I64d
& %I64u
SubmitStatusPracticeHDU
5543

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

 
length segment. And the gold sticks as segments too. There were many gold sticks with different length


 

  
and value

 

  .
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,




















 .


 
test cases follow. Each test case start with two integers,






















 
and





















 ,
represents the number of gold sticks and the length of the container stick.


 
lines follow. Each line consist of two integers,

 

 







 

 











 
and

 

 







 

 





 

 

 ,
represents the length and the value of the

 



  
gold stick.
 

Output

For each test case, output one line containing
Case #x: y, where

 
is the test case number (starting from 1) and

 
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.

题意，用一段长度的木棍装一定长度的金块，只要重心在木棍上，就可以放上去，问最多可以装下价值多少的金块，这就是背包问题，用一维背包解决，分三种状态，左右两端和全部的。我用的二维背包错了，心累啊

AC代码：

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#include <iostream>
#include <stdio.h>
#include <string>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iomanip>
#include<cmath>
#define mod 1000000007
using namespace std;
long long dp[10000][5];
long long w[10000];
int v[10000];
long long max(long long a, long long b)
{
return a > b ? a : b;
}
int main()
{
int t;
int N, L;
scanf("%d", &t);
for (int k = 1; k <= t; k++)
{
memset(dp, 0, sizeof(dp));
memset(w, 0, sizeof(w));
memset(v, 0, sizeof(v));
scanf("%d%d", &N, &L);
long long ans = 0;
for (int j = 0; j < N; j++)
{
scanf("%lld%d", &w[j], &v[j]);
ans = max(ans, v[j]);//对于第四个样例，必须的步骤
}
for (int i = 0; i < N; i++)
for (int j = L * 2; j >= w[i]; j--)//乘以2倍可以在奇数的情况下不用w/2+1
{
for (int kk = 0; kk < 3; kk++)
{
if (j >= w[i] * 2)
dp[j][kk] = max(dp[j][kk], dp[j - w[i] * 2][kk] + v[i]);
if (kk > 0)
{
dp[j][kk] = max(dp[j][kk], dp[j - w[i]][kk - 1] + v[i]);
}
ans = max(dp[j][kk], ans);
}
}
printf("Case #%d: %lld\n", k, ans);
}
}

我的错误二维代码：
#include <iostream>
#include <stdio.h>
#include <string>
#include<algorithm>
#include<cstring>
#include<string>
#include<map>
#include<iomanip>
#include<cmath>
#define mod 1000000007
using namespace std;
long long dp[1005][1005][3];
long long w[1005];
long long v[1005];
long long max(long long a, long long b)
{
return a > b ? a : b;
}
int main()
{
int t;
int N, L;
scanf("%d", &t);
for (int k = 1; k <= t; k++)
{
memset(dp, 0, sizeof(dp));
scanf("%d%d", &N, &L);
long long ans = 0;
for (int j = 1; j <= N; j++)
{
scanf("%lld%lld", &w[j], &v[j]);
ans = max(ans, v[j]);
}
for (int i = 1; i <= N; i++)
for (int j = L; w[i]&1?j>=(w[i]/2+1):j>=(w[i]/2); j--)
{
for (int kk = 0; kk < 3; kk++)
{
dp[i][j][kk] = dp[i - 1][j][kk];
if (j >= w[i])
{
dp[i][j][kk] = max(dp[i][j][kk], dp[i - 1][j - w[i]][kk] + v[i]);
}
if (kk > 0)
{
if (w[i] & 1)
dp[i][j][kk] = max(dp[i][j][kk], dp[i - 1][j - (w[i] / 2 + 1)][kk - 1] + v[i]);
else
dp[i][j][kk] = max(dp[i][j][kk], dp[i - 1][j - w[i] / 2][kk - 1] + v[i]);
}
ans = max(dp[i][j][kk], ans);
}
}
printf("Case #%d: %lld\n", k, ans);
}
}