HDU3466 Proud Merchants 排序01背包

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7030    Accepted Submission(s): 2929

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
这道题很明显是个01背包，但与之前不同的是每件物品多了一个门槛，只有你的钱比门槛高，你才能购买这件物品。
一开始我考虑的是用价值减去价格进行排序，这样买的都是赚的最多的，然后就WA了，首先应该考虑的是能不能买的问题，这就牵涉到购买第一件物品对第二件物品的购买的影响，会产生不同的价格区间
物品1：价格P1 门槛Q1 价值V1   物品2：价格P2 门槛Q2 价值V2
如果先买物品一  那么所需的最少的钱为  物品2的门槛和物品1的价格，即Q2+P1
同理，如果先买物品2，所需价钱为物品1的门槛和物品2的价格，即Q1+P2
如果第一种情况的价格比第二种高 即 Q2+P1>Q1+P2,就要选第二个，即先选2  此时Q2-P2 > Q1-P1
反之 ，就要先选第一个，此时表达式为   Q2+P1<Q1+P2  ，改写为 Q1-p1>Q2-P2
由此可以得出，要先选Q-P的值大的，即门槛减去价格所得的值大的，排序就要按这个从大到小排序
#include<stdio.h>#include<string.h>#include<algorithm>#include<stdlib.h>#include<iostream>using namespace std;struct merchine{    int price;    int low;    int val;};int cmp(merchine a,merchine b){    return     a.low-a.price < b.low-b.price;}int main(){    merchine m[505];    int inti,nums;    int dp[5050];    while(scanf("%d%d",&nums,&inti)!=EOF)    {        memset(dp,0,sizeof(dp));        memset(m,0,sizeof(m));        for(int i=1; i<=nums; i++)            scanf("%d%d%d",&m[i].price,&m[i].low,&m[i].val);        sort(m+1,m+nums+1,cmp);        for(int i=1; i<=nums; i++)        {            for(int j=inti; j>=m[i].low; j--)            {                dp[j] = max(dp[j], dp[j-m[i].price]+m[i].val);            }        }        printf("%d\n",dp[inti]);    }    return 0;}