Codeforces Round #Pi (Div. 2) D. One-Dimensional Battle Ships set区间分解

D. One-Dimensional Battle Ships
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/567/problem/D

Description

Alice and Bob love playing one-dimensional battle ships. They play on the field in the form of a line consisting of n square cells (that is, on a 1 × n table).

At the beginning of the game Alice puts k ships on the field without telling their positions to Bob. Each ship looks as a 1 × a rectangle (that is, it occupies a sequence of a consecutive squares of the field). The ships cannot intersect and even touch each other.

After that Bob makes a sequence of "shots". He names cells of the field and Alice either says that the cell is empty ("miss"), or that the cell belongs to some ship ("hit").

But here's the problem! Alice like to cheat. May be that is why she responds to each Bob's move with a "miss".

Help Bob catch Alice cheating — find Bob's first move, such that after it you can be sure that Alice cheated.

Input

The first line of the input contains three integers: n, k and a (1 ≤ n, k, a ≤ 2·105) — the size of the field, the number of the ships and the size of each ship. It is guaranteed that the n, k and a are such that you can put k ships of size a on the field, so that no two ships intersect or touch each other.

The second line contains integer m (1 ≤ m ≤ n) — the number of Bob's moves.

The third line contains m distinct integers x1, x2, ..., xm, where xi is the number of the cell where Bob made the i-th shot. The cells are numbered from left to right from 1 to n

Output

Print a single integer — the number of such Bob's first move, after which you can be sure that Alice lied. Bob's moves are numbered from1 to m in the order the were made. If the sought move doesn't exist, then print "-1".

Sample Input

11 3 3
5
4 8 6 1 11

Sample Output

3

HINT

题意

有一个一维线段，上面摆了k个船，每个船的长度都为a，然后有一个人来打，问你第几次攻击 就可以使得这个船无论怎么摆都不合法了

题解：

一个区间合并的题，对于每一次的攻击，都只会影响到这个点所在区间，然后我们更新一下新出现的两个区间里面能摆多少个船就好了

做法很多，可以set，也可以线段树，也可以离线做一下

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

set<int> S;
map<int,int> vis;
int main()
{
    int n=read(),k=read(),a=read();
    int ans=(n+1)/(a+1);
    int kiss=read();
    S.insert(0);
    S.insert(n+1);
    for(int i=0;i<kiss;i++)
    {
        int x=read();
        if(vis[x])
            continue;
        vis[x]=1;
        int c=*S.lower_bound(x);
        int d=*--S.lower_bound(x);
        S.insert(x);
        ans-=(c-d)/(a+1);
        ans+=(c-x)/(a+1)+(x-d)/(a+1);
        if(ans<k)
        {
            printf("%d
",i+1);
            return 0;
        }
    }
    cout<<"-1"<<endl;
}