hdu 3496 Watch The Movie(二维01背包)
来源:互联网 发布:nginx如何反向代理 编辑:程序博客网 时间:2024/05/24 04:34
Watch The Movie
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 6316 Accepted Submission(s): 2012
Problem Description
New semester is coming, and DuoDuo has to go to school tomorrow. She decides to have fun tonight and will be very busy after tonight. She like watch cartoon very much. So she wants her uncle to buy some movies and watch with her tonight. Her grandfather gave them L minutes to watch the cartoon. After that they have to go to sleep.
DuoDuo list N piece of movies from 1 to N. All of them are her favorite, and she wants her uncle buy for her. She give a value Vi (Vi > 0) of the N piece of movies. The higher value a movie gets shows that DuoDuo likes it more. Each movie has a time Ti to play over. If a movie DuoDuo choice to watch she won’t stop until it goes to end.
But there is a strange problem, the shop just sell M piece of movies (not less or more then), It is difficult for her uncle to make the decision. How to select M piece of movies from N piece of DVDs that DuoDuo want to get the highest value and the time they cost not more then L.
How clever you are! Please help DuoDuo’s uncle.
DuoDuo list N piece of movies from 1 to N. All of them are her favorite, and she wants her uncle buy for her. She give a value Vi (Vi > 0) of the N piece of movies. The higher value a movie gets shows that DuoDuo likes it more. Each movie has a time Ti to play over. If a movie DuoDuo choice to watch she won’t stop until it goes to end.
But there is a strange problem, the shop just sell M piece of movies (not less or more then), It is difficult for her uncle to make the decision. How to select M piece of movies from N piece of DVDs that DuoDuo want to get the highest value and the time they cost not more then L.
How clever you are! Please help DuoDuo’s uncle.
Input
The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by input data for each test case:
The first line is: N(N <= 100),M(M<=N),L(L <= 1000)
N: the number of DVD that DuoDuo want buy.
M: the number of DVD that the shop can sale.
L: the longest time that her grandfather allowed to watch.
The second line to N+1 line, each line contain two numbers. The first number is the time of the ith DVD, and the second number is the value of ith DVD that DuoDuo rated.
The first line is: N(N <= 100),M(M<=N),L(L <= 1000)
N: the number of DVD that DuoDuo want buy.
M: the number of DVD that the shop can sale.
L: the longest time that her grandfather allowed to watch.
The second line to N+1 line, each line contain two numbers. The first number is the time of the ith DVD, and the second number is the value of ith DVD that DuoDuo rated.
Output
Contain one number. (It is less then 2^31.)
The total value that DuoDuo can get tonight.
If DuoDuo can’t watch all of the movies that her uncle had bought for her, please output 0.
The total value that DuoDuo can get tonight.
If DuoDuo can’t watch all of the movies that her uncle had bought for her, please output 0.
Sample Input
13 2 1011 1001 29 1
Sample Output
3
列出n部电影 但是只能买m部 总时长不能超过l
每部电影给出时长以及看完这部电影得到的value
因为给出两个限制条件m l所以需要使用二维背包
另外价值可能为负 所以在初始化dp值的时候需要注意
#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#include <vector>#include <queue>#define MEM(a,x) memset(a,x,sizeof a)#define eps 1e-8#define MOD 10009#define MAXN 10010#define MAXM 100010#define INF 99999999#define ll __int64#define bug cout<<"here"<<endl#define fread freopen("ceshi.txt","r",stdin)#define fwrite freopen("out.txt","w",stdout)using namespace std;int Read(){ char c = getchar(); while (c < '0' || c > '9') c = getchar(); int x = 0; while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); } return x;}void Print(int a){ if(a>9) Print(a/10); putchar(a%10+'0');}int dp[110][1010];struct node{ int time,val;}no[110];int main(){ //fread; int tc; scanf("%d",&tc); while(tc--) { int n,m,l; scanf("%d%d%d",&n,&m,&l); for(int i=0;i<n;i++) scanf("%d%d",&no[i].time,&no[i].val); for(int i=0;i<=m;i++) { for(int j=0;j<=l;j++) { if(i==0) dp[i][j]=0; else dp[i][j]=-INF; } } for(int i=0;i<n;i++) { for(int j=m;j>=1;j--) { for(int k=l;k>=no[i].time;k--) { dp[j][k]=max(dp[j][k],dp[j-1][k-no[i].time]+no[i].val); } } } if(dp[m][l]<0) dp[m][l]=0; printf("%d\n",dp[m][l]); } return 0;}
0 0
- HDU 3496 Watch The Movie(二维01背包)
- hdu 3496 Watch The Movie(二维01背包)
- hdu 3496 Watch The Movie(二维01背包)
- hdu 3496 Watch The Movie(二维01背包)
- HDU-3496-Watch The Movie(二维01背包问题)
- HDU-3496-Watch The Movie(二维01背包)
- HDU 3496 Watch The Movie (二维背包+01背包)
- hdu 3496 Watch The Movie(二维背包)
- HDU 3496 Watch The Movie (二维背包)
- HDU 3496 Watch The Movie 二维背包
- HDU 3496 Watch The Movie 二维背包
- hdu 3496 Watch The Movie 二维01背包
- hdu 3496 Watch The Movie(二维01背包)
- HDU 3496 Watch The Movie(二维01背包)
- hdu 3496 Watch The Movie 二维01背包
- hdu 3496 Watch The Movie 二维01背包
- hdu 3496 Watch The Movie (二维背包)
- hdu 3496 Watch The Movie (二维费用背包)
- POI 注解方式 导入导出 excel , 只依赖于 POI包
- AndroidManifest.xml配置文件详解
- 7种经典排序算法的图解
- win8.1 64位qt5.4.0 gui程序调用控制台窗口
- TT日程管理V2.0开发系列1——经验分享
- hdu 3496 Watch The Movie(二维01背包)
- 31.modal方式弹出控制器的view
- xcode 运行错误总结
- WebRequest.GetSystemWebProxy()的效能问题
- 第11周项目3-日期时间类
- web应用的跨域访问解决方案
- 如何在Ubuntu QML应用中进行语言录音
- Linux 指令---mount(挂载命令)
- iText学习笔记