HDU-3466 Proud Merchants （和放入顺序有关的背包问题 经典题）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7117    Accepted Submission(s): 2963

Problem Description

Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more.
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?

 

Input

There are several test cases in the input.

Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.

The input terminates by end of file marker.

 

Output

For each test case, output one integer, indicating maximum value iSea could get.

 

Sample Input

2 1010 15 105 10 53 105 10 53 5 62 7 3

 

Sample Output

511

 

#include <bits/stdc++.h>using namespace std;int dp[5001];struct point{int p, q, v;bool operator < (const point& x){return q - p < x.q - x.p;}}a[5001];int main(){int n, m;while(scanf("%d %d", &n, &m) != EOF){memset(dp, 0, sizeof(dp));for(int i = 1; i <= n; ++i){scanf("%d %d %d", &a[i].p, &a[i].q, &a[i].v);}sort(a + 1, a + 1 + n);for(int i = 1; i <= n; ++i){for(int j = m; j >= a[i].q; --j){dp[j] = max(dp[j], dp[j - a[i].p] + a[i].v);}}printf("%d\n", dp[m]);}}/*题意：n个物品，m元，每个物品有价格，价值，和购买时手上最少有多少钱。问m元最多买多少价值的物品。思路：这题需要注意dp时遍历所有物品的顺序。我们考虑任意一个购买顺序，对于其中任意相邻的两个物品i，j，这两个物品的购买顺序对于后面购买的物品没有影响，因为无论前面剩下多少钱，这两个是怎样的购买顺序，最后只是消耗了pi + pj，但是会影响前面购买物品后需要剩下多少钱，假设i在j前面购买需要更少的钱，那么i在前面购买需要pi + qj,j在i前面购买需要pj + qi，pi + qj < pj + qi => qi - pi > qj - pj，差值大的在前面购买需要前面购买后剩下较少的钱，那么对于任意相邻我们都可以做这个调换。即我们先购买q-p差值大的。那么在做背包时，转移方程是dp[j] = max(dp[j], dp[j - a[i].p] + a[i].v)，实际上对于局部j，我们考虑的是最后一次购买的是a[i]，然后用dp[j - a[i].p]这个量去转移，所以我们需要先用最后购买的物品去先转移dp方程。那么就是按q-p差值从小到大更新dp。*/