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01背包 HDU3466

編輯：C++入門知識

01背包 HDU3466

因為要錢大於q的時候才可以買這樣東西，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

Sample Output

5
11

Author iSea @ WHU
Source 2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define MAXN 5000 + 100

struct 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;
}

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