HDU3577：Fast Arrangement(线段树区间更新+lazy)

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

   
   
   
   

    
    
    
    1
3 6
1 6
1 6
3 4
1 5
1 2
2 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1:
1 2 3 5 
   
   
   
   

 

 

题意：给出k，和m，有m询问，火车最多不能坐超过k个人，每个人按次序登车，能上则输出编号，不能上的就不输出

思路：可以将火车票的起点与终点看做一个区间，那么这就成了一个区间覆盖的问题，但是如果在更新状态的时候我们一直更新到叶子节点的话就会超时，说以我们要用到lazy思想

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const int L = 1000000+10;

int k,n;
int ans[L],len;
struct node
{
    int l,r,n,lazy;
} a[L<<2];

void set_up(int i)
{
    a[i].n = max(a[2*i].n,a[2*i+1].n);
}

void set_down(int i)
{
    if(a[i].lazy)
    {
        //若上面if条件不成立，则要询问它的子节点，此时增量要下传，并且要更新其本身的lazy;
        a[2*i].n += a[i].lazy;
        a[2*i].lazy+=a[i].lazy;
        a[2*i+1].n += a[i].lazy;
        a[2*i+1].lazy+=a[i].lazy;
        a[i].lazy = 0;
    }
}

void init(int l,int r,int i)
{
    a[i].l = l;
    a[i].r = r;
    a[i].n = 0;
    a[i].lazy = 0;
    if(l!=r)
    {
        int mid = (l+r)>>1;
        init(l,mid,2*i);
        init(mid+1,r,2*i+1);
    }
}

void insert(int l,int r,int i)
{
    if(a[i].l == l && a[i].r == r)
    {
        a[i].n++;
        a[i].lazy++;
        return;
    }
    set_down(i);
    int mid = (a[i].l+a[i].r)>>1;
    if(r<=mid)
        insert(l,r,2*i);
    else if(l>mid)
        insert(l,r,2*i+1);
    else
    {
        insert(l,mid,2*i);
        insert(mid+1,r,2*i+1);
    }
    set_up(i);
}

int query(int l,int r,int i)
{
    if(a[i].l == l && a[i].r == r)
        return a[i].n;
    set_down(i);
    int mid = (a[i].l+a[i].r)>>1;
    if(r<=mid)
        return query(l,r,2*i);
    else if(l>mid)
        return query(l,r,2*i+1);
    else
        return max(query(l,mid,2*i),query(mid+1,r,2*i+1));
}

int main()
{
    int t,cas = 1,i,a,b;
    for(scanf("%d",&t); t--;)
    {
        len = 0;
        scanf("%d%d",&k,&n);
        init(1,1000000,1);
        for(i = 1; i<=n; i++)
        {
            scanf("%d%d",&a,&b);
            b--;
            if(query(a,b,1)<k)
            {
                ans[len++] = i;
                insert(a,b,1);
            }
        }
        printf("Case %d:\n",cas++);
        for(i = 0; i<len; i++)
            printf("%d ",ans[i]);
        printf("\n\n");
    }

    return 0;
}