zoukankan      html  css  js  c++  java
  • Tallest Cow(线段树较易)

    题目描述

    FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.
    FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form "cow 17 sees cow 34". This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.
    For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.

    输入

    Line 1: Four space-separated integers: N, I, H and R 
    Lines 2..R+1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.

    输出

    Lines 1..N: Line i contains the maximum possible height of cow i.

    样例输入

    9 3 5 5
    1 3
    5 3
    4 3
    3 7
    9 8
    

    样例输出

    5
    4
    5
    3
    4
    4
    5
    5
    5

    区间覆盖并且不会有相交的区间。
    #include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int M=10100;
    int le[M],ri[M];
    ll N,H,I;
    ll s[M<<2],ans[M];
    struct node
    {
        int l,r;
        int sum;
    } t[M<<2];
    void pushdown(int rs)
    {
        if(s[rs])
        {
            s[rs*2]+=s[rs];
            s[rs*2+1]+=s[rs];
            s[rs]=0;
        }
    }
    void build(int l,int r,int rs)
    {
        t[rs].l=l;
        t[rs].r=r;
        s[rs]=0;
        if(l==r)
        {
            t[rs].sum=H;
            return ;
        }
        int mid=(l+r)/2;
        build(l,mid,rs*2);
        build(mid+1,r,rs*2+1);
    }
    void update(int l,int r,int rs)
    {
        if(t[rs].l>=l&&t[rs].r<=r)
        {
            s[rs]+=1;
            return ;
        }
        pushdown(rs);
        int mid=(t[rs].l+t[rs].r)/2;
        if(l<=mid)
            update(l,r,rs*2);
        if(r>mid)
            update(l,r,rs*2+1);
    }
    int query(int k,int rs)
    {
        if(t[rs].l==t[rs].r)
            return t[rs].sum-=s[rs];
        pushdown(rs);
        int mid=(t[rs].l+t[rs].r)/2;
        if(k<=mid)
            return query(k,rs*2);
        else
            return query(k,rs*2+1);
    }
    int main()
    {
        std::ios::sync_with_stdio(false);
        std::cin.tie(0);
        int i,j,k=0,m,n;
        int x,y,T,flag;
        cin>>N>>I>>H>>T;
        build(1,N,1);
        while(T--)
        {
            cin>>x>>y;
            if(x>y)
                swap(x,y);
            flag=1;
            for(i=0; i<k; i++)
                if(x==le[i]&&y==ri[i])
                {
                    flag=0;
                    break;
                }
            if(flag==0)
                continue;
            le[k]=x,ri[k++]=y;
            if(y-x==1)
                continue;
            update(x+1,y-1,1);
        }
        for(i=1; i<=N; i++)
            ans[i]=query(i,1);
        for(i=1; i<=N; i++)
            cout<<ans[i]<<endl;
        return 0;
    }
    View Code
  • 相关阅读:
    C语言寒假大作战01
    C语言I作业12—学期总结
    C语言I博客作业11
    C语言I博客作业10
    C语言I博客作业09
    C语言I博客作业08
    计算机组成与设计 复习
    概率论与数理统计 期末复习
    SPM(Software Project Management)课程感想
    Software Project Management_JUnit && Maven
  • 原文地址:https://www.cnblogs.com/nublity/p/9296618.html
Copyright © 2011-2022 走看看