hdoj Consumer 3449 （背包）好题

Consumer

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1897    Accepted Submission(s): 1010

Problem Description

FJ is going to do some shopping, and before that, he needs some boxes to carry the different kinds of stuff he is going to buy. Each box is assigned to carry some specific kinds of stuff (that is to say, if he is going to buy one of these stuff, he has to buy the box beforehand). Each kind of stuff has its own value. Now FJ only has an amount of W dollars for shopping, he intends to get the highest value with the money.

Input

The first line will contain two integers, n (the number of boxes 1 <= n <= 50), w (the amount of money FJ has, 1 <= w <= 100000) Then n lines follow. Each line contains the following number pi (the price of the ith box 1<=pi<=1000), mi (1<=mi<=10 the number goods ith box can carry), and mi pairs of numbers, the price cj (1<=cj<=100), the value vj(1<=vj<=1000000)

Output

For each test case, output the maximum value FJ can get

Sample Input

3 800300 2 30 50 25 80600 1 50 130400 3 40 70 30 40 35 60

Sample Output

210
/*题意：每种类型的物品要一个箱子中，并且每个箱子都得花钱买，问最终卖得的物品的最大价值。 给你两个数 n，w,n表示有n中类型的物品，w表示总钱数。接下来有n行，每行先输入两个数ww(箱子的价格),t(此箱中放的物品数)，    接着输入t对数，每对数有c(表示物品的价格),v（表示物品的价值） */
#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#define ll long long#define N 100010using namespace std;ll dp[N];ll a[N];int main(){int n,w;int ww,t;int i,j,k; while(scanf("%d%d",&n,&w)!=EOF){memset(dp,0,sizeof(dp));for(i=0;i<n;i++){scanf("%d%d",&ww,&t);for(j=0;j+ww<=w;j++)a[j+ww]=dp[j];for(j=1;j<=t;j++){int c,v;scanf("%d%d",&c,&v);for(k=w;k>=c+ww;k--)a[k]=max(a[k],a[k-c]+v);}for(j=ww;j<=w;j++)dp[j]=max(dp[j],a[j]);}printf("%lld\n",dp[w]);}return 0;}