Time Limit: 15000/10000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)
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 and output
| Sample Input | Sample Output |
|---|---|
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 |
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：给你一根长为m的长木板和一些小木棒，每一根小木棒有它的长度和价值，这些小木棒要放在长木板上并且每一根小木棒的重心要在长木板上
（即可以露出一半的长），问最大价值是多少。
思路：主要是dp方程要想到，用dp[i][j][k]表示处理到第i根木棒，长木板用了j的长度，有k根露在外面的最大价值，因为对于每根木棒，有三种情况，
一种是不放，一种是放在木板里面，还有一种是放在木板外面。这里注意要用滚动数组，不然会超内存。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define maxn 4005
ll l[maxn],v[maxn];
ll dp[2][maxn*2][3];
int main()
{
int n,m,i,j,T,len,cas=0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
m*=2;
for(i=1;i<=n;i++){
scanf("%lld%lld",&l[i],&v[i]);
l[i]*=2;
}
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++){
//不放
for(j=0;j<=m;j++){
dp[1][j][0]=dp[0][j][0];
dp[1][j][1]=dp[0][j][1];
dp[1][j][2]=dp[0][j][2];
}
//放里面
for(j=m;j>=l[i];j--){
dp[1][j][0]=max(dp[1][j][0],dp[0][j-l[i] ][0]+v[i]);
dp[1][j][1]=max(dp[1][j][1],dp[0][j-l[i] ][1]+v[i]);
dp[1][j][2]=max(dp[1][j][2],dp[0][j-l[i] ][2]+v[i]);
}
//放外面
for(j=m;j>=l[i]/2;j--){
dp[1][j][1]=max(dp[1][j][1],dp[0][j-l[i]/2 ][0]+v[i]);
dp[1][j][2]=max(dp[1][j][2],dp[0][j-l[i]/2 ][1]+v[i]);
}
for(j=0;j<=m;j++){
dp[0][j][0]=dp[1][j][0];
dp[0][j][1]=dp[1][j][1];
dp[0][j][2]=dp[1][j][2];
}
}
ll ans=dp[1][m][0];
ans=max(ans,dp[1][m][1]);
ans=max(ans,dp[1][m][2]);
for(i=1;i<=n;i++){
ans=max(ans,v[i]);
}
cas++;
printf("Case #%d: %lld
",cas,ans);
}
}