HDU5188：zhx and contest（类01背包）

zhx and contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 828    Accepted Submission(s): 316

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 31 4 73 64 1 86 8 101 5 22 710 4 110 2 3

 

Sample Output

78zhx is naive!

 

Source

BestCoder Round #33

题意：略。

HINT：第一组数据，第6s时开始做，第7s完成。第二组数据，第1s时做第3题，第4s时做第1题，第8s时完成。

思路·：由于限制了做题时间，需要排一下序，将先做的题放到前面再用01背包解法。

# include <stdio.h># include <string.h># include <algorithm># define MAXN 100000using namespace std;int dp[MAXN*30+1];struct node{    int t, v, l;}a[31];int cmp(node a, node b) {return a.l-a.t < b.l-b.t;}int main(){    int n, w, totalv, totalt, maxl, max_v;    while(~scanf("%d%d",&n,&w))    {        totalt = totalv = maxl = 0;        for(int i=0; i<n; ++i)        {            scanf("%d%d%d",&a[i].t, &a[i].v, &a[i].l);            totalt += a[i].t;            totalv += a[i].v;            maxl = max(maxl, a[i].l);        }        if(totalv < w)        {            puts("zhx is naive!");            continue;        }        max_v = max(totalt, maxl);        sort(a, a+n, cmp);        memset(dp, 0, sizeof(dp));        for(int i=0; i<n; ++i)            for(int j=max_v; j>=a[i].l && j>=a[i].t; --j)                dp[j] = max(dp[j], dp[j-a[i].t]+a[i].v);        for(int i=0;;++i)            if(dp[i] >= w)            {                printf("%d\n",i);                break;            }    }    return 0;}