Zepto Code Rush 2014-A. Feed with Candy（HACK）

A. Feed with Candy

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The hero of the Cut the Rope game is a little monster named Om Nom. He loves candies. And what a coincidence! He also is the hero of today's problem.

One day, Om Nom visited his friend Evan. Evan has n candies of two types (fruit drops and caramel drops), the i-th candy hangs at the height of hi centimeters above the floor of the house, its mass is mi. Om Nom wants to eat as many candies as possible. At the beginning Om Nom can make at most x centimeter high jumps. When Om Nom eats a candy of mass y, he gets stronger and the height of his jump increases by y centimeters.

What maximum number of candies can Om Nom eat if he never eats two candies of the same type in a row (Om Nom finds it too boring)?

Input

The first line contains two integers, n and x (1 ≤ n, x ≤ 2000) — the number of sweets Evan has and the initial height of Om Nom's jump.

Each of the following n lines contains three integers ti, hi, mi (0 ≤ ti ≤ 1; 1 ≤ hi, mi ≤ 2000) — the type, height and the mass of the i-th candy. If number ti equals 0, then the current candy is a caramel drop, otherwise it is a fruit drop.

Output

Print a single integer — the maximum number of candies Om Nom can eat.

Sample test(s)

input

5 3
0 2 4
1 3 1
0 8 3
0 20 10
1 5 5

output

4

Note

连续第三次被黑，尽管是一道贪心题目可是。更加注重的是，分类讨论的思想！考察思维的全面性

被黑的地方是，第一次没有依照顺序来，当前状态能吃哪就吃哪个。

这就easy漏掉一组解。

由于本题吃糖的方式仅仅能有两种：01010101... 或 10101010...分类讨论就好啊。。。SB

另外，我试了一下，究竟是不是贪心，于是删掉了。快排函数，WA在第37组，所以不用贪心能够PASS。

以后PASS了，淡定点

据本人连续三次被黑的经验，我弄明确了CFHACK数据的处理方式，每一个人不一样，HACK数据怎么增加后台数据库的？在比赛期间，HACK数据会放在你本人PASS数据库的第一组，在比赛之后，HACK数据后放在你本人数据库的最后一组数据

被黑数据

3 1
0 1 1
1 1 5
0 7 1

答案 3

分类讨论：先吃0 那么答案是3，先吃1仅仅能是2

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
const int N = 2100;
using namespace std;

struct node{
    int vis,t,h,w;
}g[N];
int cmp(const void *a,const void *b)
{
    struct node * X = (struct node *)a;
    struct node * Y = (struct node *)b;
    return   Y->w - X->w;
}
int n,x;
int main()
{
    int n,x;
    cin>>n>>x;
    int xxx = x;
    for(int i = 0;i<n;i++)
      {
          scanf("%d%d%d",&g[i].t,&g[i].h,&g[i].w);
          g[i].vis = 0;
      }
    qsort(g,n,sizeof(g[0]),cmp);
    int sum1 = 0,st=0,flag;
    for(int ll = 0;ll<n;ll++)
    {
       for (int k = 0;k < n;k++)
          {
        if (g[k].vis == 0 && x >= g[k].h)
            {
              if (st != g[k].t)
                {
                    x += g[k].w;
                    st = g[k].t;
                    g[k].vis = 1;
                    sum1++;
                    break;
                }
            }
         }
          flag = 0;
            for(int i = 0;i<n;i++)
            {
                if(g[i].vis==0 && x>=g[i].h && g[i].t!=st)
                   flag++;
            }
            if(flag==0)
                break;
    }
    for(int i = 0;i<n;i++)
        g[i].vis = 0;
    int sum2 = 0;
    st = 1;
    for(int ll = 0;ll<n;ll++)
    {
       for (int k = 0;k < n;k++)
          {
        if (g[k].vis == 0 && xxx >= g[k].h)
            {
              if (st != g[k].t)
                {
                    xxx += g[k].w;
                    st = g[k].t;
                    g[k].vis = 1;
                    sum2++;
                    break;
                }
            }
         }
          flag = 0;
            for(int i = 0;i<n;i++)
            {
                if(g[i].vis==0 && xxx>=g[i].h && g[i].t!=st)
                   flag++;
            }
            if(flag==0)
                break;
    }
    if(sum1>sum2)
        cout<<sum1<<endl;
    else
     cout<<sum2<<endl;
    return 0;
}