HDU-3577 Fast Arrangement（线段树、段变化模板）

Fast Arrangement

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3350    Accepted Submission(s): 960

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

 

Author

Louty (Special Thanks Nick Gu)

 

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

 

Recommend

zhouzeyong

 

题目大意：车上最多能装k个人，q次上下车。求哪几次是允许的。

 

解题思路：有个坑是某个人从a上车在b下车，a~b-1在车上，到b是就不在了。幸亏有人提醒。。不然觉得自己也要wa好多次。。

　　　　　就是运用lazy思想，理解后几乎就是套模板。个人觉得重点是理解pushdown、pushup函数。上一篇说过了 （http://www.cnblogs.com/WWkkk/p/7366754.html）这里就不说了。。直接上代码

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+5;
const int maxn = 4*N;
int vis[N],sum[maxn],lazy[maxn];
int k,m,a,b;

void pushdown(int num)
{
    if(lazy[num])
    {
        lazy[num*2] += lazy[num];
        lazy[num*2+1] += lazy[num];

        sum[num*2] += lazy[num];
        sum[num*2+1] += lazy[num];

        lazy[num] = 0;
    }
}

void pushup(int num)
{
    sum[num] = max(sum[num*2],sum[num*2+1]);
}

void build(int num,int l,int r)
{
    lazy[num] = 0;
    sum[num] = 0;
    if(l==r)
    {
        return ;
    }
    int mid = (l+r)/2;
    build(num*2,l,mid);
    build(num*2+1,mid+1,r);
    pushup(num);
}

void update(int num,int l,int r)
{
    if(a<=l&&b>=r)
    {
        lazy[num]++;
        sum[num] ++;
        return ;
    }
    pushdown(num);
    int mid=(l+r)/2;
    if(a<=mid)
        update(num*2,l,mid);
    if(b>mid)
        update(num*2+1,mid+1,r);
    pushup(num);
}

int query(int num,int l,int r)
{
    if(a<=l&&b>=r)
    {
        return sum[num];
    }
    int mid = (l+r)/2;
    int ans = 0;
    pushdown(num);
    if(mid>=a)
    {
        ans = max(ans,query(num*2,l,mid));
    }
    if(mid<b)
    {
        ans = max(ans,query(num*2+1,mid+1,r));
    }
    return ans;
}

int main()
{
    int T,Case=1;
    scanf("%d",&T);
    while(T--)
    {
        //memset(vis,false,sizeof(vis));
        scanf("%d %d",&k,&m);
        build(1,1,1000000);
        int n =0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&a,&b);
            b = b-1;
            int t=query(1,1,1000000);
            if(t<k)
            {
                update(1,1,1000000);
                vis[n++] = i;
            }
        }
        printf("Case %d:
",Case++);
        for(int i=0;i<n;i++)
        {
            printf("%d ",vis[i]);
        }
        printf("

");
    }
}