• Codeforces Round #244 (Div. 2) B. Prison Transfer 线段树rmq


    B. Prison Transfer

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/problemset/problem/427/B

    Description

    The prison of your city has n prisoners. As the prison can't accommodate all of them, the city mayor has decided to transfer c of the prisoners to a prison located in another city.

    For this reason, he made the n prisoners to stand in a line, with a number written on their chests. The number is the severity of the crime he/she has committed. The greater the number, the more severe his/her crime was.

    Then, the mayor told you to choose the c prisoners, who will be transferred to the other prison. He also imposed two conditions. They are,

        The chosen c prisoners has to form a contiguous segment of prisoners.
        Any of the chosen prisoner's crime level should not be greater then t. Because, that will make the prisoner a severe criminal and the mayor doesn't want to take the risk of his running away during the transfer.

    Find the number of ways you can choose the c prisoners.

    Input

    The first line of input will contain three space separated integers n (1 ≤ n ≤ 2·105), t (0 ≤ t ≤ 109) and c (1 ≤ c ≤ n). The next line will contain n space separated integers, the ith integer is the severity ith prisoner's crime. The value of crime severities will be non-negative and will not exceed 109.

    Output

    Print a single integer — the number of ways you can choose the c prisoners.

    Sample Input

    4 3 3
    2 3 1 1

    Sample Output

    2

    HINT

    题意

    给你n个人,让你选出连续c个人,要求这c个人的最大值小于t

    问你有多少种选择方法

    题解:

    正解大概双端队列吧

    我写的线段树

    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 1050005
    #define mod 10007
    #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;
    }
    //**************************************************************************************
    
    struct node
    {
        int l,r;
        int ma;
    };
    node a[maxn*4];
    int num[maxn];
    void build(int x,int l,int r)
    {
        a[x].l=l,a[x].r=r;
        if(l==r)
        {
            a[x].ma=num[l];
            return;
        }
        int mid=(l+r)>>1;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        a[x].ma=max(a[x<<1].ma,a[x<<1|1].ma);
    }
    int query(int x,int l,int r)
    {
        int L=a[x].l,R=a[x].r;
        if(l<=L&&R<=r)
            return a[x].ma;
        int mid=(a[x].l+a[x].r)>>1;
        if(r<=mid)
            return query(x<<1,l,r);
        if(l>mid)
            return query(x<<1|1,l,r);
        return max(query(x<<1,l,mid),query(x<<1|1,mid+1,r));
    }
    int main()
    {
        int n=read(),t=read(),c=read(),ans=0;
        c--;
        for(int i=1;i<=n;i++)
            num[i]=read();
        build(1,1,n);
        for(int i=1;i<=n-c;i++)
            if(query(1,i,i+c)<=t)
                ans++;
        cout<<ans<<endl;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4614476.html
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