浙江财经大学
信工学院ACM集训队

HDU 5543 Pick The Sticks

本文由 Ocrosoft 于 2016-08-11 8:24:03 发表

Pick The Sticks

Time Limit: 15000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1102    Accepted Submission(s): 350

Problem 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
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.
 

Solution

题意:给n跟木棒,一个长度为m的容器,要求将木棍放进容器使得价值最大。木棍只要重心在容器内就算放进容器,同时两根木棍不能有重叠。

思路:用f[i][j][u]表示现在处理到前i根木棍,它们在容器内占据的长度为j,且有u条木棍悬挂在外的最大价值,显然u的取值范围是[0,2] 
原长变为两倍,可以防止取中点出现小数。
那么我们有状态转移方程 
gmax(f[i][j][u],f[i-1][j-a[i].len][u]+a[i].val); 
gmax(f[i][j][u],f[i-1][j-a[i].half][u-1]+a[i].val); 
可以使用三维数组表示空间,这里可以下降为只有[j][u],第一维度的i可以省略。 
这样压缩空间的做法需要无环,要避免同一个木棍多次更新状态。 
只要改变枚举顺序,使得j降序,因为每个物品的空间都至少为1,所以不论u的顺序,这里就可以保证,一定不会重复更新。 

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 4000+10
using namespace std;
int n,m;
struct Node
{
    int half,len,value;
    bool operator<(const Node t)const
    {return len<t.len;}
} node[MAXN];
long long dp[MAXN][3];
int main()
{
    int T,cas=1;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        m*=2;//容器变为两倍
        long long ans=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&node[i].half,&node[i].value);
            node[i].len=node[i].half*2;
            ans=max(ans,(long long)node[i].value);
        }

        sort(node+1,node+1+n);//排序后去除后面半长大于容器长度的
        while(n>=1&&node[n].half>=m)n--;//全长肯定有一半以上在外面

        ms(dp);
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=node[i].half;j--)
            {
                for(int u=0;u<3;u++)
                {
                    if(j>=node[i].len)//能放下整根木棍
                        dp[j][u]=max(dp[j][u],dp[j-node[i].len][u]+node[i].value);
                    if(u)//放一半,另一半放在外部
                        dp[j][u]=max(dp[j][u],dp[j-node[i].half][u-1]+node[i].value);
                }
            }
        }
        //有可能放一根会价值更大
        printf("Case #%d: %lld\n",cas++,max(ans,dp[m][2]));
    }
    return 0;
}

 

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