POJ Topic 2392 Space Elevator (multiple backpack)

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9042		Accepted: 4296

Description

The cows is going to space! They plan to achieve orbit by building a sort of space elevator:a giant Tower of blocks. They has K (1 <= k <=) different types of blocks with which to build the tower. Each block of type I has a height h_i (1 <= h_i <=) and is available in quantity c_i (1 <= c_i <= 10). Due to possible damage caused by cosmic rays, no part of a block of type I can exceed a maximum altitude a_i (1 <= a_i <= 40000).

Help the cows build the tallest space elevator possible by stacking blocks on top of each of the other according to the rules.

Input

* Line 1: A single integer, K

* Lines 2..k+1:each line contains three space-separated integers:h_i, a_i, and c_i. Line i+1 describes block type I.

Output

* Line 1: A single integer H, the maximum height of a tower that can be built

Sample Input

37 40 35 23 82 52 6

Sample Output

48

Hint

OUTPUT DETAILS:

From the Bottom:3 blocks of type 2, below 3 of type 1, below 6 of type 3. Stacking 4 blocks of type 2 and 3 of type 1 are not legal, since the top of the last type 1 block would exceed height 40.

Source

Usaco 2005 March Gold

AC Code

#include <stdio.h> #include <string.h> #include <stdlib.h> #define INF 1<<30#define Max (A, B) >B?A:B) #define MIN (A, b) (a>b?a:b) struct s{int h,num,mh;} B[440];int dp[40040];int cmp (const void *a,const void *b) {return (* (struct S *) a). mh-(* (struct S *) b). MH; int N;int Main () {while (scanf ("%d", &n)!=eof) {int i,j,k;for (i=0;i<n;i++) {scanf ("%d%d%d", &b[i].h,&b[i ].mh,&b[i].num);} Qsort (B,n,sizeof (b[0]), CMP), memset (Dp,0,sizeof (DP)), for (i=0;i<n;i++) {for (k=1;k<=b[i].num;k++) {for (J=b[i] . mh;j>=b[i].h;j--) {Dp[j]=max (dp[j],dp[j-b[i].h]+b[i].h);}}} int Ans=0;for (i=0;i<=b[n-1].mh;i++) {Ans=max (ans,dp[i]);} printf ("%d\n", ans);}}

POJ Topic 2392 Space Elevator (multiple backpack)