hdu 3577 Fast Arrangement（区间更新）

最新推荐文章于 2019-11-04 21:48:39 发布

理想主义文艺青年

最新推荐文章于 2019-11-04 21:48:39 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/u011699990/article/details/48919239

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针对中国铁路旅客众多的问题，本文介绍了一种使用线段树算法实现的高效火车票查询系统设计方案。该系统确保每趟列车不超过最大乘客容量，并遵循先到先得的原则分配座位。

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Fast Arrangement

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2745 Accepted Submission(s): 766

Problem Description

Chinese always have the railway tickets problem because of its' huge amount of passangers and stations. Now goverment need you to develop a new tickets query system.
One train can just take k passangers. And each passanger can just buy one ticket from station a to station b. Each train cannot take more passangers any time. The one who buy the ticket earlier which can be sold will always get the ticket.

Input

The input contains servel test cases. The first line is the case number. In each test case:
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 )
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query.
Huge Input, scanf recommanded.

Output

For each test case, output three lines:
Output the case number in the first line.
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number.
Output a blank line after each test case.

Sample Input

Sample Output

Case 1:
1 2 3 5

火车买票火车最多搭载k个人有q个人去买票买从a到b的票问这个人能不能买到票

线段树的想法用每个叶子节点模拟每一个车站中火车上人数的情况

注意的是买从a到b的票该人是在a到b-1站中占着火车中的一个人数

在更新之前要先查询这个区间的火车还能不能载人

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 1000010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}
int kk,q;

struct node
{
    int l,r;
    int num,lazy;
};
node no[MAXN*4];
void build(int l,int r,int k)
{
    if(l==r)
    {
        no[k].l=l;
        no[k].r=r;
        no[k].num=kk;
        no[k].lazy=0;
        return;
    }
    int mid=(l+r)/2;
    no[k].l=l;
    no[k].r=r;
    no[k].num=kk;
    no[k].lazy=0;
    build(l,mid,k*2);
    build(mid+1,r,k*2+1);
}

void pushdown(int k)
{
    if(no[k].lazy)
    {
        no[k*2].lazy+=no[k].lazy;
        no[k*2+1].lazy+=no[k].lazy;
        no[k*2].num-=no[k].lazy;
        no[k*2+1].num-=no[k].lazy;
        no[k].lazy=0;
        return;
    }
}

void update(int l,int r,int k)
{
    if(l<=no[k].l&&r>=no[k].r)
    {
        no[k].lazy++;
        no[k].num--;
        return;
    }
    pushdown(k);
    int mid=(no[k].l+no[k].r)/2;
    if(r<=mid) update(l,r,k*2);
    else if(l>mid) update(l,r,k*2+1);
    else
    {
        update(l,mid,k*2);
        update(mid+1,r,k*2+1);
    }
    no[k].num=min(no[k*2].num,no[k*2+1].num);
}

int query(int l,int r,int k)
{
    if(no[k].l>=l&&no[k].r<=r)
    {
        return no[k].num;
    }
    pushdown(k);
    int mid=(no[k].l+no[k].r)/2;
    if(r<=mid) return query(l,r,k*2);
    else if(l>mid) return query(l,r,k*2+1);
    else
    {
        return min(query(l,mid,k*2),query(mid+1,r,k*2+1));
    }
}

int main()
{
//    fread;
    int cs=1;
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        scanf("%d%d",&kk,&q);
        build(1,1000000,1);
        int flag=0;
        printf("Case %d:\n",cs++);
        for(int i=1;i<=q;i++)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            r--;
            if(query(l,r,1)>0)
            {
//                if(flag)
//                    printf(" ");
//                else flag=1;
                printf("%d ",i);
                update(l,r,1);
            }
        }
        printf("\n\n");
    }
    return 0;
}