A. Feed with Candy
time limit per test
memory limit per test
input
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.
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
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
n and x (1 ≤ n, x ≤ 2000)
n lines contains three integers ti, hi, mi (0 ≤ ti ≤ 1; 1 ≤ hi, mi — the type, height and the mass of the i-th candy. If number ti
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
原本代码交上了 谁知道被一外国小伙Hack了。没办法誰让自己的代码没有人家的更完美啊，经过修改AC的代码终于写出来了。代码如下：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct node
{
int a,bp,c,d;
}x[2010];
int cmp (const void *a,const void *b)
{
struct node * aa = (struct node *)a;
struct node * bb = (struct node *)b;
return bb->c - aa->c;
}
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
int count1 = 0,count2 = 0,h = 1;
for(i=0;i<2010;i++)
{
x[i].d = 0;
}
for(i=0;i<n;i++)
{
scanf("%d%d%d",&x[i].a,&x[i].bp,&x[i].c);
}
qsort(x,n,sizeof(x[0]),cmp);
int pp = 1;
int tt = m;
while(h == 1)
{
for(i=0;i<n;i++)
{
if(tt>=x[i].bp && pp!=x[i].a && x[i].d == 0)
{
tt = tt + x[i].c;
x[i].d = 1;
pp = x[i].a;
count1++;
break;
}
}
if(i == n)
{
h == 0;
break;
}
}
h == 1;
pp = 0;
for(i=0;i<2010;i++)
{
x[i].d = 0;
}
while(h == 1)
{
for(i=0;i<n;i++)
{
if(m>=x[i].bp && pp!=x[i].a && x[i].d == 0)
{
m = m + x[i].c;
x[i].d = 1;
pp = x[i].a;
count2++;
break;
}
}
if(i == n)
{
h = 0;
break;
}
}
if(count1>count2)
{
printf("%d\n",count1);
}
else
{
printf("%d\n",count2);
}
}
return 0;
}
下面是被Hack的代码：
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct node
{
int a,bp,c,d;
}x[2010];
int cmp (const void *a,const void *b)
{
struct node * aa = (struct node *)a;
struct node * bb = (struct node *)b;
return bb->c - aa->c;
}
int main()
{
int n,m,i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
int count = 0,h = 1;
for(i=0;i<2010;i++)
{
x[i].d = 0;
}
for(i=0;i<n;i++)
{
scanf("%d%d%d",&x[i].a,&x[i].bp,&x[i].c);
}
qsort(x,n,sizeof(x[0]),cmp);
int k = 0;
int pp = 11;
while(h == 1)
{
k++;
if(k == 1)
{
for(i=0;i<n;i++)
{
if(m>=x[i].bp)
{
m = m + x[i].c;
x[i].d = 1;
pp = x[i].a;
count++;
break;
}
}
}
else
{
for(i=0;i<n;i++)
{
if(m>=x[i].bp && pp!=x[i].a && x[i].d == 0)
{
m = m + x[i].c;
x[i].d = 1;
pp = x[i].a;
count++;
break;
}
}
if(i == n)
{
h = 0;
break;
}
}
}
printf("%d\n",count);
}
return 0;
}
Note
4 candies is to eat them in the order: 1, 5, 3, 2. Let's assume the following scenario:
- 3. He can reach candies 1 and 2. Let's assume that he eats candy 1. As the mass of this candy equals 4, the height of his jump will rise to 3 + 4 = 7.
- 2 and 5. Let's assume that he eats candy 5. Then the height of his jump will be 7 + 5 = 12.
- 2 and 3. He won't eat candy 2 as its type matches the type of the previously eaten candy. Om Nom eats candy 3, the height of his jump is 12 + 3 = 15.
- 2, the height of his jump is 15 + 1 = 16. He cannot reach candy 4.