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

Memory Limit: 256 MB

題目連接

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

?

題意

有一個(gè)一維線段，上面擺了k個(gè)船，每個(gè)船的長(zhǎng)度都為a，然后有一個(gè)人來打，問你第幾次攻擊 就可以使得這個(gè)船無論怎么擺都不合法了

題解：

一個(gè)區(qū)間合并的題，對(duì)于每一次的攻擊，都只會(huì)影響到這個(gè)點(diǎn)所在區(qū)間，然后我們更新一下新出現(xiàn)的兩個(gè)區(qū)間里面能擺多少個(gè)船就好了

做法很多，可以set，也可以線段樹，也可以離線做一下

代碼

#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\n",i+1);return 0;}}cout<<"-1"<<endl; }

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轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/4706305.html

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