2015CCPC南阳场 D - Pick The Sticks

D - Pick The Sticks

Time Limit:10000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 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：给你一个容器的体积，再给n个物品，往容器里面放东西，只要物品的重心在容器里面就可以了，所以物品的总体积有可能大于容器的体积，这是这道题的难点，否则就是普通的背包了。我们可以这样想，再开多一维数组来记录边缘没有完全放进容器里的物品个数，因为只有两端，所以最多也就两个物品分别在两端没有完全在容器里面。那么这个问题就化简为简单的三维背包问题了。还有两个问题就是，容器里面能只放一个任意大的物品，因为只要把物品放在中间，怎么都是平衡的，所以要考虑只放一个的情况。

还有另外一个问题就是，只把重心放在容器里面把体积的一半放进去，这是会出现小数的，所以我们应该把全部体积乘上2再计算。下面给代码。

#include<iostream>
#include<stack>
#include<cstring>
#include<map>
#include<string>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
int main(){
	int t;
	scanf("%d", &t);
	for (int tcase = 1; tcase <= t;tcase++){
		int n, l;
		scanf("%d%d", &n, &l);
		int a[1005], b[1005];
		LL dp[4010][2][3];
		LL maxnum = 0;
		for (int i = 0; i < n; i++){
			scanf("%d%d", &a[i], &b[i]);
			a[i] =a[i]<<1;
			maxnum = max((LL)b[i], maxnum);
		}
		memset(dp, 0, sizeof(dp));
		int now = 1;
		for (int i = 0; i < n; i++){
			now = now ^ 1;
			for (int j = 1; j <= l<<1; j++){
				for (int k = 0; k < 3; k++){
					dp[j][now][k] = dp[j][now ^ 1][k];
					if (j - a[i] >= 0){
						dp[j][now][k] = max(dp[j][now][k], dp[j - a[i]][now ^ 1][k]+b[i]);
					}
					if (k&&(j - (a[i]>>1))  >= 0){
						dp[j][now][k] = max(dp[j][now][k], dp[j - (a[i] >>1)][now ^ 1][k-1]+b[i]);
					}
				}
			}
		}
		maxnum = max(maxnum, dp[l<<1][now][0]);
		maxnum = max(maxnum, dp[l<<1][now][1]);
		maxnum = max(maxnum, dp[l<<1][now][2]);
		printf("Case #%d: %lld\n",tcase, maxnum);
	}
}

思路二：

用dp[i][M][k] 表示用前i种物品放在长度为M的大木块上，两端放了k个小木棒的做法。

所以dp[i][j][k]=max   dp[i-1][j-a[i].length][k] + a[i].value

       dp[i][j][k]=max   dp[i-1][j-a[i].length/2][k-1] + a[i].value

但是这样写会爆内存，那就降维吧。方法跟二维01背包降维一样

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1005;
typedef long long LL;
LL dp[4*maxn][4],maxvalue;
int T,N,M;
struct node
{
	LL value,length;
}a[maxn];
void init() 
{
	memset(dp,0,sizeof(dp));
}
void input()//输入 
{
	scanf("%d%d",&N,&M);
	maxvalue=0;
	M*=2;
	for(int i=1;i<=N;i++)
	{
		scanf("%lld%lld",&a[i].length,&a[i].value);
		a[i].length=a[i].length*2;
		maxvalue=max(maxvalue,a[i].value);
	}
}
int main()
{
	scanf("%d",&T);
	for(int tcase=1;tcase<=T;tcase++)
	{
		init();
		input();
		for(int i=1;i<=N;i++)//dp[i][j]  把前i件物品放入容量为j的背包的最大收益
		{
			for(int j=M;j>=a[i].length/2;j--)
			{
				for(int k=0;k<3;k++) 
				{
					if(j-a[i].length>=0)dp[j][k]=max(dp[j][k],dp[j-a[i].length][k]+a[i].value);
					if(k>0)
					{
						if(j-a[i].length/2>=0)dp[j][k]=max(dp[j][k],dp[j-a[i].length/2][k-1]+a[i].value);
					}
				}
			}
		}
		printf("Case #%d: ",tcase);
		LL ans=max(max(dp[M][0],dp[M][1]),max(dp[M][2],maxvalue));
		cout<<ans<<endl;
	}
}