poj 2392 Space Elevator(多重背包)

Space Elevator

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7618 Accepted: 3591

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

USACO 2005 March Gold

题意：

给你n种砖。告诉你每种砖的高度，数目。和最多承受高度(也就是最大能处的海拔)。问你用这些砖能建多高。

思路：

先按最大承受海拔排序然后多重背包。f[i]表示。1高度i能达到。0为不能。

由于最大承受海拔的原因更新范围有限必须排序。不排序就不是最优的。

详细见代码：

#include<algorithm>#include<iostream>#include<string.h>#include<sstream>#include<stdio.h>#include<math.h>#include<vector>#include<string>#include<queue>#include<set>#include<map>//#pragma comment(linker,"/STACK:1024000000,1024000000")using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-8;const double PI=acos(-1.0);const int maxn=40010;typedef __int64 ll;int num[maxn],f[maxn],ans;struct node{    int h,a,c;} blo[450];bool cmp(node a,node b){    return a.a<b.a;}void pack(int val,int mou,int lim)//wei==val时的特殊情况{    for(int i=val;i<=lim;i++)    {        if(f[i])            num[i]=0;        else if(f[i-val])//如果f[i-val]为1有两种可能。1,还没有使用物品i        {                //2,已经使用了物品i。可知1的num为0。2的num洽为使用了的个数            if(i<2*val)//这范围的还没初始化            {                num[i]=1;                f[i]=1;                ans=max(ans,i);            }            else if(num[i-val]<mou)            {                num[i]=num[i-val]+1;                f[i]=1;                ans=max(ans,i);            }        }    }}int main(){    int n,i;    while(~scanf("%d",&n))    {        ans=0;        for(i=0;i<n;i++)            scanf("%d%d%d",&blo[i].h,&blo[i].a,&blo[i].c);        sort(blo,blo+n,cmp);        memset(f,0,sizeof f);        f[0]=1;        for(i=0;i<n;i++)            pack(blo[i].h,blo[i].c,blo[i].a);        printf("%d\n",ans);    }    return 0;}