UESTC 1227 & POJ 3667 Hotel

非常细腻的线段树题目啊,后来还是有个细节写错了,查了一个晚上。。就不分析了。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <utility>
#include <cstdlib>
using namespace std;
#define N 80011

struct node
{
    int ls,rs,ms;
    int pos;
    int mark;   // 0: unsure  1: all-empty  2: all-full
}tree[4*N];

int n,m;

void build(int l,int r,int rt)
{
    tree[rt].ls = tree[rt].rs = tree[rt].ms = r-l+1;
    tree[rt].pos = l;
    if(l == r)
    {
        return;
    }
    int mid = (l+r)/2;
    build(l,mid,2*rt);
    build(mid+1,r,2*rt+1);
}

int if_all_empty(int l,int r,int rt)
{
    if(tree[rt].ls == r-l+1)
        return 1;
    return 0;
}

void update(int l,int r,int rt)
{
    if(!tree[rt].mark)   
        return;
    if(tree[rt].mark == 1)  //全空,则下传给左右子树
    {
        int len = r-l+1;
        tree[2*rt].ls = tree[2*rt].rs = tree[2*rt].ms = (len+1)/2;
        tree[2*rt].pos = l;
        tree[2*rt+1].ls = tree[2*rt+1].rs = tree[2*rt+1].ms = len/2;
        tree[2*rt+1].pos = (l+r)/2+1;
        tree[2*rt].mark = tree[2*rt+1].mark = 1;
    }
    else if(tree[rt].mark == 2)  //全满,则下传给左右子树
    {
        tree[2*rt].ls = tree[2*rt].rs = tree[2*rt].ms = 0;
        tree[2*rt].pos = l;
        tree[2*rt+1].ls = tree[2*rt+1].rs = tree[2*rt+1].ms = 0;
        tree[2*rt+1].pos = (l+r)/2+1;
        tree[2*rt].mark = tree[2*rt+1].mark = 2;
    }
    tree[rt].mark = 0;   // not "== 0"
}

int query(int l,int r,int dis,int rt)
{
    update(l,r,rt);
    if(tree[rt].ms<dis)
        return 0;
    if(tree[rt].ms == dis)
        return tree[rt].pos;
    int mid = (l+r)/2;
    if(tree[2*rt].ms>=dis)
        return query(l,mid,dis,2*rt);
    if(tree[2*rt].rs + tree[2*rt+1].ls>=dis)
        return mid - tree[2*rt].rs + 1;
    return query(mid+1,r,dis,2*rt+1);
}

void in_out(int l,int r,int aa,int bb,int flag,int rt)   //flag == 1: insert   else  quit
{
    if(aa>r||bb<l)
        return;
    if(aa<=l&&bb>=r)
    {
        if(flag == 1)   //如果当前要入住
        {
            tree[rt].ls = tree[rt].rs = tree[rt].ms = 0;
            tree[rt].pos = l;
            tree[rt].mark = 2;
        }
        else  //如果当前要退房
        {
            tree[rt].ls = tree[rt].rs = tree[rt].ms = r-l+1;
            tree[rt].pos = l;
            tree[rt].mark = 1;
        }
        return;
    }
    update(l,r,rt);
    int mid = (l+r)/2;
    in_out(l,mid,aa,bb,flag,2*rt);     
    in_out(mid+1,r,aa,bb,flag,2*rt+1);

    tree[rt].ls = tree[2*rt].ls;
    if(if_all_empty(l,mid,2*rt))
        tree[rt].ls += tree[2*rt+1].ls;
    tree[rt].rs = tree[2*rt+1].rs;
    if(if_all_empty(mid+1,r,2*rt+1))
        tree[rt].rs += tree[2*rt].rs;

    tree[rt].ms = max(tree[2*rt].rs+tree[2*rt+1].ls,max(tree[2*rt].ms,tree[2*rt+1].ms));

    if(tree[rt].ms == tree[2*rt].ms)   //如果当前区间最大空房数等于左子树最大空房数
        tree[rt].pos = tree[2*rt].pos;  //则起点置为左子树的起点

    else if(tree[rt].ms == tree[2*rt].rs + tree[2*rt+1].ls)  //同理
        tree[rt].pos = mid - tree[2*rt].rs + 1;

    else
        tree[rt].pos = tree[2*rt+1].pos;
}

int main()
{
    scanf("%d%d",&n,&m);
    memset(tree,0,sizeof(tree));
    build(1,n,1);
    int i,flag;
    int x,dis;
    for(i=0;i<m;i++)
    {
        scanf("%d",&flag);
        if(flag == 1)
        {
            scanf("%d",&dis);
            int ans = query(1,n,dis,1);
            printf("%d
",ans);
            if(ans)
                in_out(1,n,ans,ans+dis-1,1,1);
        }
        else if(flag == 2)
        {
            scanf("%d%d",&x,&dis);
            in_out(1,n,x,x+dis-1,2,1);
        }
    }
    return 0;
}
View Code

作者:whatbeg
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原文地址:https://www.cnblogs.com/whatbeg/p/3488318.html