01背包 HDU3466

因为要钱大于q的时候才可以买这样东西，所以要先排序

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2765    Accepted Submission(s): 1146

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

 

#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include <limits.h>#include <ctype.h>#include <string.h>#include <string>#include <math.h>#include <algorithm>#include <iostream>#include <queue>#include <deque>#include <vector>#include <set>#include <map>using namespace std;#define MAXN 5000 + 100struct node{    int p,q,v;}a[MAXN];int dp[MAXN];bool cmp(node a,node b){    return a.q-a.p < b.q-b.p;}int Max(int a,int b){    return a>b?a:b;}int main(){    int n,m,i,j;    while(~scanf("%d%d",&n,&m)){        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++){            scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);        }        sort(a+1,a+n+1,cmp);        for(i=1;i<=n;i++){            for(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]);    }    return 0;}