time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Do you know the story about the three musketeers? Anyway, you must help them now.
Richelimakieu is a cardinal in the city of Bearis. He found three brave warriors and called them the three musketeers. Athos has strength a, Borthos strength b, and Caramis has strength c.
The year 2015 is almost over and there are still n criminals to be defeated. The i-th criminal has strength ti. It’s hard to defeat strong criminals — maybe musketeers will have to fight together to achieve it.
Richelimakieu will coordinate musketeers’ actions. In each hour each musketeer can either do nothing or be assigned to one criminal. Two or three musketeers can be assigned to the same criminal and then their strengths are summed up. A criminal can be defeated in exactly one hour (also if two or three musketeers fight him). Richelimakieu can’t allow the situation where a criminal has strength bigger than the sum of strengths of musketeers fighting him — a criminal would win then!
In other words, there are three ways to defeat a criminal.
A musketeer of the strength x in one hour can defeat a criminal of the strength not greater than x. So, for example Athos in one hour can defeat criminal i only if ti ≤ a.
Two musketeers can fight together and in one hour defeat a criminal of the strength not greater than the sum of strengths of these two musketeers. So, for example Athos and Caramis in one hour can defeat criminal i only if ti ≤ a + c. Note that the third remaining musketeer can either do nothing or fight some other criminal.
Similarly, all three musketeers can fight together and in one hour defeat a criminal of the strength not greater than the sum of musketeers’ strengths, i.e. ti ≤ a + b + c.
Richelimakieu doesn’t want musketeers to fight during the New Year’s Eve. Thus, he must coordinate their actions in order to minimize the number of hours till all criminals will be defeated.
Find the minimum number of hours to defeat all criminals. If musketeers can’t defeat them all then print “-1” (without the quotes) instead.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of criminals.
The second line contains three integers a, b and c (1 ≤ a, b, c ≤ 108) — strengths of musketeers.
The third line contains n integers t1, t2, …, tn (1 ≤ ti ≤ 108) — strengths of criminals.
Output
Print one line with the answer.
If it’s impossible to defeat all criminals, print “-1” (without the quotes). Otherwise, print the minimum number of hours the three musketeers will spend on defeating all criminals.
Examples
input
5
10 20 30
1 1 1 1 50
output
2
input
5
10 20 30
1 1 1 1 51
output
3
input
7
30 20 10
34 19 50 33 88 15 20
output
-1
input
6
10 5 10
10 9 5 25 20 5
output
3
Note
In the first sample Athos has strength 10, Borthos 20, and Caramis 30. They can defeat all criminals in two hours:
Borthos and Caramis should together fight a criminal with strength 50. In the same hour Athos can fight one of four criminals with strength 1.
There are three criminals left, each with strength 1. Each musketeer can fight one criminal in the second hour.
In the second sample all three musketeers must together fight a criminal with strength 51. It takes one hour. In the second hour they can fight separately, each with one criminal. In the third hour one criminal is left and any of musketeers can fight him.
【题解】
贪心；
用平衡树来找小于等于某个能力值的抵抗力最大的敌人,并删掉它;
用代码来讲;(以一小时为单位，查看一小时，这3个人能做些什么);
//a[1]<a[2]<a[3]
//work(a)会删去平衡树中小于等于a且最大的数字;
int mx = get_max(root);
if (mx<=a[1])//如果抵抗力最大的敌人都比能力值最小的那个人小。则三个人每次都能打3个敌人
{
ans--;
ans +=(n+2)/3;
break;
}
else
if (mx<=a[2])//如果mx<a[1]<=a[2];
{//则分别去打就好;
work(a[1]);
work(a[2]);
work(a[3]);
}
else
if (mx <=a[3])//如果a[1]<a[2]<mx<=a[3]
{//如果a[1]和a[2]每个人都能打一个敌人(两个),则让他们俩分别打；如果两人总共只能打一只或不能打,尝试a[1]+a[2]合起来打一只会更好;
qianqu = 0;
ask_before(root,a[1]);
bool flag = 1;
if (qianqu!=0)
{
int temp1 = qianqu;
de_lete(root,qianqu);
qianqu = 0;
ask_before(root,a[2]);
if (qianqu!=0)
{
de_lete(root,qianqu);
n-=2;
work(a[3]);
}
else
{
insert(root,temp1);
flag = 0;
}
}
else
flag = 0;
if (!flag)
{
work(a[1]+a[2]);
work(a[3]);
}
}
else
if (mx<=a[1]+a[2])//小于最小的两个，就a[1]+a[2]合体然后a[3]分别打;
{
work(a[1]+a[2]);
work(a[3]);
}
else
if (mx <= a[1]+a[3])//同理
{
work(a[1]+a[3]);
work(a[2]);
}
else
if (mx <= a[2]+a[3])
{
work(a[2]+a[3]);
work(a[1]);
}
else//想等就全部合起来打;
work(a[1]+a[2]+a[3]);完整代码↓↓↓
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <string>
#define lson L,m,rt<<1
#define rson m+1,R,rt<<1|1
#define LL long long
using namespace std;
const int MAXN = 6000990;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
int n,root=0,l[MAXN],r[MAXN],key[MAXN],size[MAXN],totn = 0,cf[MAXN],a[4],qianqu;
int t[MAXN];
int get_max(int t)
{
int temp = r[t];
while (temp)
{
t = r[t];
temp = r[t];
}
return key[t];
}
void right_rotation(int &t)
{
int k = l[t];
l[t] = r[k];
r[k] = t;
size[k] = size[t];
size[t] = size[l[t]]+size[r[t]]+1;
t = k;
}
void left_rotation(int &t)
{
int k = r[t];
r[t] = l[k];
l[k] = t;
size[k] = size[t];
size[t] = size[r[t]]+size[l[t]]+1;
t = k;
}
void maintain(int &t,bool flag)
{
if (flag)
{
if (size[l[l[t]]] > size[r[t]])
right_rotation(t);
else
if (size[r[l[t]]] > size[r[t]])
{
left_rotation(l[t]);
right_rotation(t);
}
else
return;//如果都不是的话就结束递归
}
else
{
if (size[r[r[t]]] > size[l[t]])
{
left_rotation(t);
}
else
if (size[l[r[t]]] > size[l[t]])
{
right_rotation(r[t]);
left_rotation(t);
}
else
return;
}
maintain(l[t],true);
maintain(r[t],false);
maintain(t,true);
maintain(t,false);
}
void insert(int &t,int data)
{
if (t==0)
{
t=++totn;
l[t] = r[t] = 0;
size[t] = 1;
key[t] = data;
}
else
{
size[t]++;
if (data<key[t])
insert(l[t],data);
else
insert(r[t],data);
maintain(t,data<key[t]);
}
}
void ask_before(int t, int x)
{//询问前趋是啥
if (!t)
return;
if (key[t] <= x)
{
qianqu = max(qianqu, key[t]);
ask_before(r[t], x);
}
else
if (x < key[t])
ask_before(l[t], x);
}
void de_lete(int &t,int data) //这是删除平衡树中某个节点的过程;
{
size[t]--;//因为这个节点肯定在以t为根的树下面。所以删除后，它的大小会递减。
if (key[t] == data)//如果找到了要删除的元素。
{
if (l[t] ==0 && r[t] == 0)//如果它没有左子树或右子树
t = 0;//那么就直接把这个节点置空
else
if (l[t] == 0 && r[t]!=0)//如果左子树为空，右子树不为空
t = r[t];//就把这个节点去掉，用右子树来接在下面。
else
if (l[t] != 0 && r[t] == 0)//如果左子树不为空，右子树为空
t = l[t];//则直接把这个节点去掉，把左子树接在下面。
else
if (l[t]!=0 && r[t]!=0)
{
int temp = r[t];
while (l[temp]) temp = l[temp];
key[t] = key[temp];
de_lete(r[t],key[temp]);
}
}
else
if (data<key[t])
de_lete(l[t],data);
else
de_lete(r[t],data);
}
void work(int x)
{
if (!root)
return;
qianqu = 0;
ask_before(root,x);
if (qianqu!=0)
{
de_lete(root,qianqu);
n--;
}
}
int main()
{
//freopen("F:\rush.txt","r",stdin);
//printf("%d
",6*sizeof(size)/1024/1024);
// return 0;
scanf("%d",&n);
for (int i = 1;i <= 3;i++)
scanf("%d",&a[i]);
sort(a+1,a+1+3);
for (int i = 1;i <= n;i++)
scanf("%d",&t[i]),insert(root,t[i]);
if (get_max(root)>a[1]+a[2]+a[3])
{
puts("-1");
return 0;
}
int ans = 0;
while (root && n)
{
ans++;
int mx = get_max(root);
if (mx<=a[1])
{
ans--;
ans +=(n+2)/3;
break;
}
else
if (mx<=a[2])
{
work(a[1]);
work(a[2]);
work(a[3]);
}
else
if (mx <=a[3])
{
qianqu = 0;
ask_before(root,a[1]);
bool flag = 1;
if (qianqu!=0)
{
int temp1 = qianqu;
de_lete(root,qianqu);
qianqu = 0;
ask_before(root,a[2]);
if (qianqu!=0)
{
de_lete(root,qianqu);
n-=2;
work(a[3]);
}
else
{
insert(root,temp1);
flag = 0;
}
}
else
flag = 0;
if (!flag)
{
work(a[1]+a[2]);
work(a[3]);
}
}
else
if (mx<=a[1]+a[2])
{
work(a[1]+a[2]);
work(a[3]);
}
else
if (mx <= a[1]+a[3])
{
work(a[1]+a[3]);
work(a[2]);
}
else
if (mx <= a[2]+a[3])
{
work(a[2]+a[3]);
work(a[1]);
}
else
work(a[1]+a[2]+a[3]);
}
printf("%d
",ans);
return 0;
}