hdoj3466Proud Merchants(排序+01背包)

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

 

题意：有n个商品，没个商品包括3个信息，商人给出的价格p，买每件商品至少需要的钱数q，以及商品本来的价格w。按(p-q)从小到大排序，然后就是01背包.ps：感觉此题有点乱

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[5500];struct node{int p,q,w;}arr[550];bool cmp(node a,node b){return a.q-a.p<b.q-b.p;}int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;i++){scanf("%d%d%d",&arr[i].p,&arr[i].q,&arr[i].w);}memset(dp,0,sizeof(dp));sort(arr+1,arr+n+1,cmp);for(i=1;i<=n;i++){for(j=m;j>=arr[i].q;j--){dp[j]=max(dp[j],dp[j-arr[i].p]+arr[i].w);}}printf("%d\n",dp[m]);}return 0;}