POJ 2392 Space Elevator

原题：

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.

题目大意：有一头奶牛要上太空，他有很多种石头，每种石头的高度是hi，但是不能放到ai之上的高度，并且这种石头有ci个

将这些石头叠加起来，问能够达到的最高高度。 
解题思路：首先对数据进行升序排序，这样才是一个标准的多重背包的问题

为什么要排序？
因为只有这样才能得到最优解,如果一开始就是高的在前面，那么后面有低的却不能选到，就直接选高的去了。这样是不能达到最优解的
使f[i]的状态标记，是否可以达到这个高度
这样能够达到取f[i]中i的最大值即可。

这里要注意max赋初值的时候要赋值为0，不能为-1，因为答案有可能为0 
做的时候不知道看数据的大小，以为对它进行二进制优化转为01背包会快一点，结果优化之后时间跟空间都变大了，而且把自己给搞糊涂了，每个物体的所在最大高度限制不知道怎么处理了，高了很长一段时间才给想明白，真是自己给自己找事。二进制转化之后的空间为12324 KB，时间为313ms，网上大神写的简单代码空间只要484kb，时间只要63ms。
自己写的长长的代码： 弄巧成拙

#include<iostream>#include<stdio.h>#include<string.h>#include <algorithm>#define MAX 1000000using namespace std;int n,m;int h[405],c[405],w[405],a[405];//a为价值，w为花费，c为数量 h为高度struct block{    int h,c,w,a;} num[405];int Count;int value[MAX];int size[MAX];int height[MAX];int dp[100005];bool cmp(block a,block b){    if(a.h<b.h)        return true;    return false;}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        Count=1;        memset(value,0,sizeof(value));        memset(size,0,sizeof(size));        memset(height,0,sizeof(height));        int mheight=0;        for(int i=1; i<=n; i++)        {            scanf("%d%d%d",&num[i].a,&num[i].h,&num[i].c);            if(num[i].h>mheight)                mheight=num[i].h;        }        sort(num+1,num+n+1,cmp);        for(int i=1; i<=n; i++)        {            for (int k=1; k<=num[i].c; k<<=1)            {                value[Count] = k*num[i].a;                height[Count]=num[i].h;                size[Count++] = k*num[i].a;                num[i].c -= k;            }            if (num[i].c>0)            {                value[Count] = num[i].c*num[i].a;                height[Count]=num[i].h;                size[Count++] = num[i].c*num[i].a;            }        }        memset(dp,0,sizeof(dp));        for(int i=1; i<=Count; i++)            for(int j=mheight; j>=size[i]; j--)                if(j<=height[i])                    dp[j]=max(dp[j],dp[j-size[i]]+value[i]);        int Max=0;        for(int k=0;k<=mheight; k++)        {            if(dp[k]>Max)            {                Max=dp[k];            }        }        printf("%d\n",Max);    }    return 0;}

大家就当看多重背包的二进制优化吧
网上大神的简易代码

#include <iostream>#include <algorithm>using namespace std;#define MAXV 410#define MAXM 40010typedef struct {int h,a,c;}Blocks;Blocks v[MAXV];int cmp(Blocks x,Blocks y){return x.a<y.a;}int f[MAXM],user[MAXM];int main(){int i,t,j,max;while(~scanf("%d",&t)){for(i=1;i<=t;i++){scanf("%d%d%d",&v[i].h,&v[i].a,&v[i].c);}sort(v+1,v+t+1,cmp);memset(f,0,sizeof(f));f[0]=1;max=0;//赋值为0，不能为-1for(i=1;i<=t;i++){memset(user,0,sizeof(user));for (j=v[i].h;j<=v[i].a;j++){if(!f[j] && f[j-v[i].h] && user[j-v[i].h]+1<=v[i].c){f[j]=1;user[j]=user[j-v[i].h]+1;//用于记录该石头用过的个数if(j>max) max=j;}}}printf("%d\n",max);}return 0;}