STL(set_pair)运用 CF#Pi D. One-Dimensional Battle Ships

D. One-Dimensional Battle Ships

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., x_m, where x_i 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 from 1 to m in the order the were made. If the sought move doesn't exist, then print "-1".

这题我一开始用裸数组写的，写的很烦。对于我这种代码能力不太好能学习不少，有两份代码，第二份非常简洁，某位大神的。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<set>
#include<algorithm>
#include<utility>
using namespace std;
#define PII pair<int,int>
#define MP make_pair
#define N 200005

int k,len,m,n;
int pri[N],num,ans;
set<PII> myset;
set<PII>::iterator it;

int main(){
    int i,j,k;
    while(~scanf("%d",&n)){
        myset.clear();
        scanf("%d %d %d",&k,&len,&m);
        PII tp = MP(1,n);
        myset.insert(tp);
        ans=-1;
        num=(tp.second-tp.first+2) / (len+1); //计算船的最大数量 记住两个船之间不能接触然后加2是因为为第一条前补1
        bool flag=true;

        for(i=0;i<m;i++){
            scanf("%d",&pri[i]);
            if(flag){
                it =myset.lower_bound(MP(pri[i],n + 1)); //找到所在区间
                it--;
                PII p =*it;
                myset.erase(it);
                num-= (p.second-p.first+2) / (len + 1);  //现将这个区间可能存在的船数剪掉
                if(p.first <= pri[i]-1){
                    PII tp= MP(p.first , pri[i]-1);
                    num+= (tp.second-tp.first+2) / (len+1);
                    myset.insert(tp);
                }
                if(pri[i]+1 <= p.second){
                    PII tp= MP(pri[i]+1 , p.second);
                    num+= (tp.second-tp.first+2) / (len+1);
                    myset.insert(tp);
                }
                if(num<k){     //如果小于K就不成立了
                    ans=i+1;
                    flag=false;
                }
            }
        }

        printf("%d\n",ans);
    }
    return 0;
}
//令一种解法

#include <bits/stdc++.h>

using namespace std;

int N, K, W, M;
int A[200000];
int B[200000];

int solve(int X)
{
    for(int i=0; i<=X; i++)
        B[i]=A[i];
    sort(B, B+X+1);
    int ret=0;
    if(B[0]>W)
        ret+=1+(B[0]-1-W)/(W+1);
    if(N-B[X]>=W)
        ret+=1+(N-B[X]-W)/(W+1);
    for(int i=1; i<=X; i++)
        if(B[i]-B[i-1]-1>=W)
            ret+=1+(B[i]-B[i-1]-1-W)/(W+1);
    return ret;
}

int main()
{
    scanf("%d%d%d", &N, &K, &W);
    scanf("%d", &M);
    for(int i=0; i<M; i++)
        scanf("%d", A+i);
    int lo=0, mid, hi=M;
    while(lo<hi)
    {
        mid=(lo+hi)/2;
        if(solve(mid)<K)
            hi=mid;
        else
            lo=mid+1;
    }
    if(lo==M)
        printf("-1\n");
    else
        printf("%d\n", lo+1);
    return 0;
}