描述：

　　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.?

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.? 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.? 代碼：

　　第一個背包是電影累加的時間，第二個背包是電影的數(shù)目。

　　值得注意的是，電影的數(shù)目背包要求解必須是正好為m，即“恰好被裝滿”，根據(jù)背包九講，如果要求背包恰好裝滿，那么此時只有容量為0 的背包可以在什么也不裝且價值為0 的情況下被“恰好裝滿”，其它容量的背包均沒有合法的解，屬于未定義的狀態(tài)，應該被賦值為-∞ 了。

　　最后如果值為負數(shù)，說明無解。

#include<stdio.h> #include<string.h> #include<iostream> #include<stdlib.h> #include <math.h> using namespace std; #define N 105 #define M 1005 #define MAX 999999int main(){int tc;int n,m,l;//n為物品個數(shù)，m為可買的數(shù)量最大值，l為時間累計最大值int t[N],v[N],dp[M][N];//dp一維為時間，二維為數(shù)量scanf("%d",&tc);while( tc-- ){scanf("%d%d%d",&n,&m,&l);for( int i=1;i<=n;i++ )scanf("%d%d",&t[i],&v[i]);for( int i=0;i<=l;i++ ){for( int j=0;j<=m;j++ ){if( j==0 )dp[i][j]=0;elsedp[i][j]=-MAX;}}for( int i=1;i<=n;i++ ){for( int j=l;j>=t[i];j-- ){//時間for( int k=m;k>=1;k-- ){//數(shù)量dp[j][k]=max(dp[j][k],dp[j-t[i]][k-1]+v[i]);}}}if( dp[l][m]<0 ) dp[l][m]=0;//為負代表沒有解printf("%d\n",dp[l][m]);}system("pause");return 0; }

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轉(zhuǎn)載于:https://www.cnblogs.com/lucio_yz/p/4722799.html

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