poj 2392 Space Elevator

Space Elevator

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12817		Accepted: 6092

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

 

题意：

用物块堆塔,每一个物块都有不同的高度以及数量，并且每一种物块能够堆放的最大高度也是有限制的。求利用这些物块能堆的最高高度。

思路：完全背包,可以利用单调队列.因为不同类型的物块能堆放的最大高度不同，最大高度限制越低的物块应当越早堆放越好，故一开始要以最大高度限制对不同种类的物块从小到大排序。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define N_MAX 400000+20
struct tower {
    int w, v, m, a;//m是数量,a是当前积木能堆积的最高高度
    tower() {}
    bool operator < (const tower &b) const{
        return a < b.a;
    }
}T[400+20];
int n;
int dp[N_MAX];
int deq[N_MAX],deqv[N_MAX];

void solve() {
    //memset(dp, 0, sizeof(dp)); memset(deq, 0, sizeof(deq)); memset(deqv,0,sizeof(deqv));
    for (int i = 0; i < n;i++) {
        for (int a = 0; a < T[i].w;a++) {
            int s = 0, t = 0;
            for (int j = 0; j*T[i].w + a <= T[i].a; j++) {
                int val = dp[j*T[i].w + a] - j*T[i].v;
                while (s < t&&deqv[t - 1] <= val)t--;
                deq[t] = j;
                deqv[t++] = val;
                dp[j*T[i].w + a] = deqv[s] + j*T[i].v;
                if (deq[s] == j - T[i].m) s++;
            }
        }
  }
}
int main() {
        scanf("%d",&n);
        for (int i = 0; i < n;i++) {
            scanf("%d%d%d",&T[i].w,&T[i].a,&T[i].m);
            T[i].v =T[i].w;
        }
        sort(T,T+n);
        solve();
        int max_height = *max_element(dp,dp+N_MAX-19);;
        printf("%d\n",max_height);
    
    return 0;
}