hdu 3535 AreYouBusy 分组背包
2024-09-29 22:07:46
AreYouBusy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Problem Description
Happy New Term!
As having become a junior, xiaoA recognizes that there is not much time for her to AC problems, because there are some other things for her to do, which makes her nearly mad.
What's more, her boss tells her that for some sets of duties, she must choose at least one job to do, but for some sets of things, she can only choose at most one to do, which is meaningless to the boss. And for others, she can do of her will. We just define the things that she can choose as "jobs". A job takes time , and gives xiaoA some points of happiness (which means that she is always willing to do the jobs).So can you choose the best sets of them to give her the maximum points of happiness and also to be a good junior(which means that she should follow the boss's advice)?
As having become a junior, xiaoA recognizes that there is not much time for her to AC problems, because there are some other things for her to do, which makes her nearly mad.
What's more, her boss tells her that for some sets of duties, she must choose at least one job to do, but for some sets of things, she can only choose at most one to do, which is meaningless to the boss. And for others, she can do of her will. We just define the things that she can choose as "jobs". A job takes time , and gives xiaoA some points of happiness (which means that she is always willing to do the jobs).So can you choose the best sets of them to give her the maximum points of happiness and also to be a good junior(which means that she should follow the boss's advice)?
Input
There are several test cases, each test case begins with two integers n and T (0<=n,T<=100) , n sets of jobs for you to choose and T minutes for her to do them. Follows are n sets of description, each of which starts with two integers m and s (0<m<=100), there are m jobs in this set , and the set type is s, (0 stands for the sets that should choose at least 1 job to do, 1 for the sets that should choose at most 1 , and 2 for the one you can choose freely).then m pairs of integers ci,gi follows (0<=ci,gi<=100), means the ith job cost ci minutes to finish and gi points of happiness can be gained by finishing it. One job can be done only once.
Output
One line for each test case contains the maximum points of happiness we can choose from all jobs .if she can’t finish what her boss want, just output -1 .
Sample Input
3 3
2 1
2 5
3 8
2 0
1 0
2 1
3 2
4 3
2 1
1 1
2 1
2 5
3 8
2 0
1 0
2 1
3 2
4 3
2 1
1 1
3 4
2 1
2 5
3 8
2 0
1 1
2 8
3 2
4 4
2 1
1 1
1 1
1 0
2 1
5 3
2 0
1 0
2 1
2 0
2 2
1 1
2 0
3 2
2 1
2 1
1 5
2 8
3 2
3 8
4 9
5 10
Sample Output
5
13
-1
-1
13
-1
-1
Author
hphp
Source
题意:n个分组,s=0最少取一个,s=1最多取一个,s=2,01背包;
思路:分组背包;
s=0,hdu 3033;
s=1,hdu 1712;
s=2,01背包;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e3+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int dp[N][N];
int v[N],w[N];
// 0 分组最多一个
// 1 分组最少一个
// 2 01背包
int main()
{
int n,T;
while(~scanf("%d%d",&n,&T))
{
memset(dp[],,sizeof(dp[]));
for(int i=;i<=n;i++)
{
int x,flag;
scanf("%d%d",&x,&flag);
for(int i=;i<=x;i++)
scanf("%d%d",&v[i],&w[i]);
if(flag==)
{
for(int t=;t<=T;t++)
dp[i][t]=-inf;
for(int t=;t<=x;t++)
{
for(int j=T;j>=v[t];j--)
{
dp[i][j]=max(dp[i][j],dp[i][j-v[t]]+w[t]);
dp[i][j]=max(dp[i][j],dp[i-][j-v[t]]+w[t]);
}
}
}
else if(flag==)
{
for(int t=;t<=T;t++)
dp[i][t]=dp[i-][t];
for(int t=;t<=x;t++)
{
for(int j=T;j>=v[t];j--)
{
dp[i][j]=max(dp[i][j],dp[i-][j-v[t]]+w[t]);
}
}
}
else
{
for(int t=;t<=T;t++)
dp[i][t]=dp[i-][t];
for(int t=;t<=x;t++)
{
for(int j=T;j>=v[t];j--)
{
dp[i][j]=max(dp[i][j],dp[i][j-v[t]]+w[t]);
dp[i][j]=max(dp[i][j],dp[i-][j-v[t]]+w[t]);
}
}
}
}
if(dp[n][T]>=)
printf("%d\n",dp[n][T]);
else
printf("-1\n");
}
return ;
}
最新文章
- 苹果官方发布,iPhone 6 &; Plus 设计素材
- linux -- 基于mysql tomcat 部署web项目
- 如何在 Windows Azure 的虚拟机 ubuntu 上面安装和配置 openVPN(一)
- runnable:在线IDE+代码片段分享
- Andorid Binder进程间通信---Binder本地对象,实体对象,引用对象,代理对象的引用计数
- Windows Phone开发(30):图形
- poj 2288 Islands and Bridges
- angr进阶(5)内存操作
- jmeter笔记(2)--组件介绍
- WPF DataGrid分页功能实现代码
- .net 服务因为GC时遇到的问题和解决办法
- 分部类,分部方法 - 修饰符partial
- PJ初赛复习日记
- MySQL各个版本区别及问题总结
- random库的常见用法
- WP-PostViews使用
- Inferred type &#39;S&#39; for type parameter &#39;S&#39; is not within its bound; should extend
- [synergy]两台机器公用键盘鼠标
- 【多视图几何】TUM 课程 第3章 透视投影
- 【nodejs】使用response输出中文但页面中文乱码的处置