2014
04-09

# Watch The Movie

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.

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 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.

1
3 2 10
11 100
1 2
9 1

3

#include<stdio.h>
#include<string.h>
#define inf 999999
int dp[102][1002],t[102],v[102];
int max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,m,i,j,k,L,T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&L);
for(i=0;i<n;i++)
scanf("%d%d",&t[i],&v[i]);
for(i=0;i<=m;i++)
for(j=0;j<=L;j++)
{
if(i==0)
dp[i][j]=0;
else
dp[i][j]=-inf; //注意初始化为负无穷，价值可能是负值
}
for(i=0;i<n;i++)
for(j=m;j>=1;j--)
for(k=L;k>=t[i];k--)
dp[j][k]=max(dp[j][k],dp[j-1][k-t[i]]+v[i]);
if(dp[m][L]<0)
dp[m][L]=0;
printf("%d\n",dp[m][L]);
}
return 0;
}


1. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

2. 因为是要把从字符串s的start位到当前位在hash中重置，修改提交后能accept，但是不修改居然也能accept