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
这道题和 poj 1742 有点类似，只是这道题要先按每个block的最大高度进行排序，这样做的目的是为了获得最优解，想想怎么证明？
然后将dp数组初始化为-1，dp=-1表示取不到，dp[i]>=0表示取到i的时候还能剩下多少个
最后结果就是从最大高度开始寻找dp[i]!=-1的i的值，直接输出即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<map> 6 #include<set> 7 using namespace std; 8 #define N 406 9 #define M 40006 10 struct Node{ 11 int h,a,c; 12 }block[N]; 13 int dp[M*N]; 14 bool cmp(Node a,Node b){ 15 return a.a<b.a; 16 } 17 int main() 18 { 19 int n; 20 while(scanf("%d",&n)==1){ 21 for(int i=0;i<n;i++){ 22 scanf("%d%d%d",&block[i].h,&block[i].a,&block[i].c); 23 } 24 sort(block,block+n,cmp); 25 26 memset(dp,-1,sizeof(dp)); 27 dp[0]=0; 28 for(int i=0;i<n;i++){ 29 for(int j=0;j<=M;j++){ 30 if(dp[j]>=0){ 31 dp[j]=block[i].c; 32 } 33 else if(j<block[i].h || dp[j-block[i].h]<=0){ 34 dp[j]=-1; 35 } 36 else if(j>block[i].a){ 37 dp[j]=-1; 38 } 39 else{ 40 dp[j]=dp[j-block[i].h]-1; 41 } 42 } 43 } 44 for(int i=M;i>=0;i--){ 45 if(dp[i]!=-1){ 46 printf("%d ",i); 47 break; 48 } 49 } 50 51 52 53 54 55 } 56 return 0; 57 }