HDU3466 Proud Merchants [背包]
Proud Merchants
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7536 Accepted Submission(s): 3144
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?
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.
10 15 10
5 10 5
3 10
5 10 5
3 5 6
2 7 3
分析:一开始考虑直接01背包,将限制条件直接换成j>=q[i],但是发现这种思路是错的。显然物品的顺序会对其产生影响,那么就要考虑如何排序。然后试了几种排序发现都不对,还是K_lord和five20大佬讲过以后才明白。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<iomanip>
using namespace std;
const int N=5e3+;
int n,m,dp[N];
struct Node{int p,q,v;}a[N];
bool cmp(Node x,Node y)
{return x.q-x.p<y.q-y.p;}
int main()
{
ios::sync_with_stdio(false);
while(cin>>n>>m){
for(int i=;i<=n;i++)
cin>>a[i].p>>a[i].q>>a[i].v;
memset(dp,,sizeof(dp));
sort(a+,a+n+,cmp);
for(int i=;i<=n;i++)
for(int j=m;j>=a[i].q;j--)
dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v);
cout<<dp[m]<<"\n";}
return ;
}
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