POJ2392Space Elevator（多重背包）

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8701 Accepted: 4135

Description

The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K <= 400) different types of blocks with which to build the tower. Each block of type i has height h_i (1 <= h_i <= 100) 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 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 is not legal, since the top of the last type 1 block would exceed height 40.

Source

大意:有K种block去建塔，每种每个都有一个高度H，用了当前的block塔的高度不能超出a,和每种的数量。求塔最高能建多高。

分析：这题就是一个多重背包，但有一点变动，必须先以a从小到大排序，因为如果先用了充许塔最高的block,而那种block的h很小，个数很少，更新自然就小,那么接下来小的就充许塔高越建越小，这样就不是我们所求的塔高了。如果先用a小的类形，就有变动更大的余地。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;struct node{    int h,a,c;}block[405];bool cmp(node a,node b){    return a.a<b.a;}int main(){    int dp[40005],k,maxa;    while(scanf("%d",&k)>0)    {        memset(dp,0,sizeof(dp));        dp[0]=1;        maxa=0;        for(int i=0;i<k;i++)            scanf("%d%d%d",&block[i].h,&block[i].a,&block[i].c);        sort(block,block+k,cmp);        int h,a,c;        for(int i=0;i<k;i++)        {            h=block[i].h;            a=block[i].a;            c=block[i].c;            if(maxa<a)                 maxa=a;            if(h*c>=a)            {                for(int i=h;i<=a;i++)                    if(dp[i-h])                     dp[i]=1;            }            else            {                int m=1;                while(c>m)                {                    for(int i=a;i>=h*m; i--)                        if(dp[i-h*m])                         dp[i]=1;                    c-=m; m*=2;                }                if(c)                    for(int i=a;i>=h*c; i--)                        if(dp[i-h*c])                         dp[i]=1;            }        }        for( ; maxa>=0; maxa--)        if(dp[maxa])        {            printf("%d\n",maxa); break;        }    }}