HDU 3466

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7161    Accepted Submission(s): 2973

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

 

Author

iSea @ WHU

 

题意：

背包，加一个限制。

POINT：

q-p 从小到大排。死活半天想不出。

#include <stdio.h>#include <map>#include<iostream>#include <math.h>#include <algorithm>#include <string.h>#include <vector>#include <queue>#include <set>using namespace std;typedef long long  LL;const int maxn = 600+55;int dp[6000];struct node{    int p,v,q;    double val;}a[maxn];bool cmd(node a ,node b){    return a.q-a.p<b.q-b.p;}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);            a[i].val=1.0*a[i].v/a[i].p;        }        sort(a+1,a+1+n,cmd);        for(int i=1;i<=n;i++){            for(int j=m;j>=a[i].q&&j>=a[i].p;j--){                dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v);            }        }        printf("%d\n",dp[m]);    }    return 0;}