POJ2392-Space Elevator-多重背包

原题链接
Space Elevator
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11392 Accepted: 5433
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

3
7 40 3
5 23 8
2 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

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 40000 + 10;typedef struct blocks{    int h,a,c;}blocks;blocks a[410];int dp[maxn];bool cmp(blocks x,blocks y){    return x.a < y.a;}int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i].h,&a[i].a,&a[i].c);    sort(a+1,a+1+n,cmp);//对最大高度进行排序再按照多重背包的方法处理就好了    memset(dp,-1,sizeof(dp));    for(int i=1;i<=n;i++){        dp[0] = a[i].c;        for(int j=1;j<=a[i].a;j++){            if(dp[j] >= 0) dp[j] = a[i].c;            else if(j < a[i].h || dp[j-a[i].h] <= 0) dp[j] = -1;            else dp[j] = dp[j-a[i].h] - 1;        }    }    int loc=0;//注意这里一定要初始化为0，因为有可能给出的石头一个都不能用来搭电梯    for(int i = a[n].a;i>0;i--){        if(dp[i] >= 0){            loc = i;            break;        }    }    cout << loc << endl;    return 0;}//下面是标准的滚动数组的形式，实际上本题可以更简化为1维dp如上// #include <cstdio>// #include <cstring>// #include <iostream>// #include <algorithm>// using namespace std;// const int maxn = 40000 + 10;// typedef struct blocks// {//  int h,a,c;// }blocks;// blocks a[410];// int dp[2][maxn];// bool cmp(blocks x,blocks y){//  return x.a < y.a;// }// int main(){//  int n;//  scanf("%d",&n);//  for(int i=1;i<=n;i++) scanf("%d%d%d",&a[i].h,&a[i].a,&a[i].c);//  sort(a+1,a+1+n,cmp);//对最大高度进行排序再按照多重背包的方法处理就好了//  memset(dp,-1,sizeof(dp));//  for(int i=1;i<=n;i++){//      dp[i&1][0] = a[i].c;//      for(int j=1;j<=a[i].a;j++){//          if(dp[(i-1)&1][j] >= 0) dp[i&1][j] = a[i].c;//          else if(j < a[i].h || dp[i&1][j-a[i].h] <= 0) dp[i&1][j] = -1;//          else dp[i&1][j] = dp[i&1][j-a[i].h] - 1;//      }//  }//  int loc=0;//注意这里一定要初始化为0，因为有可能给出的石头一个都不能用来搭电梯//  for(int i = maxn - 1;i>0;i--){//      if(dp[n&1][i] >= 0){//          loc = i;//          break;//      }//  }//  cout << loc << endl;//  return 0;// }