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.
代码：
第一个背包是电影累加的时间，第二个背包是电影的数目。
值得注意的是，电影的数目背包要求解必须是正好为m，即“恰好被装满”，根据背包九讲，如果要求背包恰好装满，那么此时只有容量为0 的背包可以在什么也不装且价值为0 的情况下被“恰好装满”，其它容量的背包均没有合法的解，属于未定义的状态，应该被赋值为-∞ 了。
最后如果值为负数，说明无解。
#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 999999
int main(){
int tc;
int n,m,l;//n为物品个数，m为可买的数量最大值，l为时间累计最大值
int t[N],v[N],dp[M][N];//dp一维为时间，二维为数量
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;
else
dp[i][j]=-MAX;
}
}
for( int i=1;i<=n;i++ ){
for( int j=l;j>=t[i];j-- ){//时间
for( int k=m;k>=1;k-- ){//数量
dp[j][k]=max(dp[j][k],dp[j-t[i]][k-1]+v[i]);
}
}
}
if( dp[l][m]<0 ) dp[l][m]=0;//为负代表没有解
printf("%d\n",dp[l][m]);
}
system("pause");
return 0;
}