HDOJ 5188 zhx and contest 贪婪+01背包

zhx and contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 455    Accepted Submission(s): 158

Problem Description

As one of the most powerful brushes in the world, zhx usually takes part in all kinds of contests.
One day, zhx takes part in an contest. He found the contest very easy for him.
There are n problems in the contest. He knows that he can solve the ith problem in ti units of time and he can get vi points.
As he is too powerful, the administrator is watching him. If he finishes the ith problem before time li, he will be considered to cheat.
zhx doesn't really want to solve all these boring problems. He only wants to get no less than w points. You are supposed to tell him the minimal time he needs to spend while not being considered to cheat, or he is not able to get enough points. 
Note that zhx can solve only one problem at the same time. And if he starts, he would keep working on it until it is solved. And then he submits his code in no time.

 

Input

Multiply test cases(less than 50). Seek EOF as the end of the file.
For each test, there are two integers n and w separated by a space. (1≤n≤30, 0≤w≤109)
Then come n lines which contain three integers ti,vi,li. (1≤ti,li≤105,1≤vi≤109)

 

Output

For each test case, output a single line indicating the answer. If zhx is able to get enough points, output the minimal time it takes. Otherwise, output a single line saying "zhx is naive!" (Do not output quotation marks).

 

Sample Input

1 3
1 4 7
3 6
4 1 8
6 8 10
1 5 2
2 7
10 4 1
10 2 3

 

Sample Output

7
8
zhx is naive!

 

Source

BestCoder Round #33

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年03月19日 星期四 10时26分15秒
File Name     :HDOJ5188.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=200100;

int n,w;
int sumtime,sumv;
int dp[maxn*30];

struct PB
{
	int t,v,l;
}pb[50];

bool cmp(PB a,PB b)
{
	return a.l-a.t<b.l-b.t;
}


int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	while(scanf("%d%d",&n,&w)!=EOF)
	{
		sumv=0;
		for(int i=0;i<n;i++)
		{
			int x,y,z;
			scanf("%d%d%d",&x,&y,&z);
			pb[i]=(PB){x,y,z};
			sumv+=y;
		}
		if(sumv<w)
		{
			puts("zhx is naive!");
			continue;
		}

		sort(pb,pb+n,cmp);
		memset(dp,0,sizeof(dp));
		for(int i=0;i<n;i++)
			for(int j=maxn;j>=max(pb[i].t,pb[i].l);j--)
				dp[j]=max(dp[j],dp[j-pb[i].t]+pb[i].v);

		int ans=0;
		for(int i=0;i<maxn;i++)
			if(dp[i]>=w) { ans=i; break; }

		printf("%d
",ans);
	}
    
    return 0;
}

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