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.
题意:有n种石头,每个石头的高度hi,允许用的最高高度为ai,数量为ci,求能组合的最高高度。
思路:先给这些石头排个序,最高高度小的在前面,这样才能使得高度最大,然后再按多重背包来做,这样就可以了。
#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct node{
int ci,hi,ai;
bool operator < (const node &a)const
{
return ai<a.ai;
}
}num[500];
bool cmp(node a,node b)
{
return a.ai<b.ai;
}
int f[40010];
int dp[40010];
int main() {
int n;
// memset(dp,-1,sizeof dp);
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d%d",&num[i].hi,&num[i].ai,&num[i].ci);
sort(num,num+n);
dp[0]=1;
for(int i=0;i<n;i++)
{
for(int j=1;j<=num[i].ci;j++)
{
for(int k=num[i].ai;k>=num[i].hi;k--)
dp[k]|=dp[k-num[i].hi];
}
}
for(int i=40000;i>=0;i--)
if(dp[i]==1){
printf("%d\n",i);break;
}
return 0;
}
上面的方法类似01背包来解决 ,但两种方法都是将ai从小到大排好序,然后枚举数量来找到最优解
#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct node{
int ci,hi,ai;
bool operator < (const node &a)const
{
return ai<a.ai;
}
}num[500];
bool cmp(node a,node b)
{
return a.ai<b.ai;
}
int f[40010];
int dp[40010];
int main() {
int n;
memset(dp,-1,sizeof dp);
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d%d",&num[i].hi,&num[i].ai,&num[i].ci);
sort(num,num+n);
dp[0]=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<=num[i].ai;j++)
{
if(dp[j]>=0)
dp[j]=num[i].ci;
else{
int t=j-num[i].hi;
if(t>=0&&dp[t]>0)
dp[j]=dp[t]-1;
}
}
}
for(int i=40000;i>=0;i--)
if(dp[i]>=0){
printf("%d\n",i);break;
}
return 0;
}