hdoj Watch The Movie 3496 (二维01背包)好题
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Watch The Movie
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 6951 Accepted Submission(s): 2197
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//题意:每个测试样例输入三个数n,m,l;n表示多多喜欢看的光盘种类为n种,m表示商店只有m中光盘种类,l表示多多看光盘的总时长。接下来是n行输入,每行两个数分别表示一种光盘的时间和价值。问多多在时间l内最多能看多少价值的光盘。//思路:二维01背包。动态转移方程为:dp[j][k]=max(dp[j][k],dp[j-1][k-q[i].t]+q[i].v);dp[j][k]表示在时长为k的时间内看j种光盘的最大价值。#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define INF 0x3f3f3f3fusing namespace std;struct zz{int t;int v;}q[1010];int cmp(zz a,zz b){return a.t<b.t;}int dp[110][1010];int main(){int t,n,m,l;int i,j,k;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&l);for(i=0;i<=m;i++){for(j=0;j<=l;j++)dp[i][j]=-INF;}for(i=0;i<=l;i++)dp[0][i]=0;for(i=0;i<n;i++)scanf("%d%d",&q[i].t,&q[i].v);for(i=0;i<n;i++){for(j=m;j>=1;j--){for(k=l;k>=q[i].t;k--){dp[j][k]=max(dp[j][k],dp[j-1][k-q[i].t]+q[i].v);}}}if(dp[m][l]>0)printf("%d\n",dp[m][l]);elseprintf("0\n");}return 0;}
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