Zhenya moves from parents

Time limit: 1.0 second
Memory limit: 64 MB
Zhenya moved from his parents’ home to study in other city. He didn’t take any cash with him, he only took his father’s credit card with zero balance on it. Zhenya succeeds in studies at the University and sometimes makes a little money on the side as a Maths tutor. As he makes his own money he spends only it, and when it is over he uses the credit card. Every time he gets or spends money, he sends a letter to his father, where he puts the following two things.
  1. The date when it took place
  2. The sum of earned or spent money
Every time receiving a letter from Zhenya, his father calculates the debt on the credit card at the moment. But here a problem arises. The point is that Russian Post delivers letters in an order different to the one they were sent in.
For example, in the first Zhenya’s letter the father read that on September 10 Zhenya spent one thousand rubles. He thought that his son had used the credit card, and now the debt is one thousand rubles. However the next day came a letter with the information that on September 9 Zhenya earned five hundred rubles. It means that half of the money he spent on September 10 was his own, and the debt on the credit card is just five hundred rubles.
Help Zhenya’s father with his account management.

Input

The first line contains an integer n which is the number of Zhenya’s letters (1 ≤ n ≤ 100 000). These letters are listed in the next n lines. Description of each letter consists of the amount of money Zhenya spent or earned (in the form -c or +c accordingly, where c is an integer, 1 ≤ c ≤ 50 000) followed by both date and time when it took place (in the form of dd.MM hh:mm). All dates belong to the same year, which is not leap (i. e. there are 365 days in it). Any two letters contain either different dates or different time. The letters are listed in the order the father received them.

Output

After each received letter output what Zhenya’s father thinks the amount of the debt on the credit card is.

Sample

input output
5
-1000 10.09 21:00
+500 09.09 14:00
+1000 02.09 00:00
-1000 17.09 21:00
+500 18.09 13:00
-1000
-500
0
-500
-500

分析:当前的盈亏始终对之后的盈亏有影响,且要维护整体最小值,所以用线段树解决;

   注意要用long long;

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+;
const int dis[][]={{,},{-,},{,-},{,}};
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p%mod;p=p*p%mod;q>>=;}return f;}
int n,m,k,t,c[maxn];
struct Node
{
ll Min, lazy;
} T[maxn<<];
void PushUp(int rt)
{
T[rt].Min = min(T[rt<<].Min, T[rt<<|].Min);
}
void PushDown(int L, int R, int rt)
{
int mid = (L + R) >> ;
ll t = T[rt].lazy;
T[rt<<].Min += t;
T[rt<<|].Min += t;
T[rt<<].lazy += t;
T[rt<<|].lazy += t;
T[rt].lazy = ;
}
void Update(int l, int r, int v, int L, int R, int rt)
{
if(l==L && r==R)
{
T[rt].lazy += v;
T[rt].Min += v;
return ;
}
int mid = (L + R) >> ;
if(T[rt].lazy) PushDown(L, R, rt);
if(r <= mid) Update(l, r, v, Lson);
else if(l > mid) Update(l, r, v, Rson);
else
{
Update(l, mid, v, Lson);
Update(mid+, r, v, Rson);
}
PushUp(rt);
}
ll Query(int l, int r, int L, int R, int rt)
{
if(l==L && r== R)
{ return T[rt].Min;
}
int mid = (L + R) >> ;
if(T[rt].lazy) PushDown(L, R, rt);
if(r <= mid) return Query(l, r, Lson);
else if(l > mid) return Query(l, r, Rson);
return min(Query(l, mid, Lson), Query(mid + , r, Rson));
}
struct node
{
int x,y;
}a[maxn];
int main()
{
int i,j;
scanf("%d",&n);
rep(i,,n)
{
int b,p,d,e;
scanf("%d",&a[i].x);
scanf("%d.%d %d:%d",&b,&p,&d,&e);
a[i].y=(b+p*)**+d*+e;
c[i]=a[i].y;
}
sort(c+,c+n+);
rep(i,,n)a[i].y=lower_bound(c+,c+n+,a[i].y)-c;
rep(i,,n)
{
Update(a[i].y,n,a[i].x,,n,);
printf("%lld\n",min(0LL,Query(,n,,n,)));
}
//system("pause");
return ;
}

ural2014 Zhenya moves from parents的更多相关文章

  1. Gym 100507C Zhenya moves from parents (线段树)

    Zhenya moves from parents 题目链接: http://acm.hust.edu.cn/vjudge/contest/126546#problem/C Description Z ...

  2. ural 2014 Zhenya moves from parents

    2014. Zhenya moves from parents Time limit: 1.0 secondMemory limit: 64 MB Zhenya moved from his pare ...

  3. zhenya moves from parents

    Zhenya moved from his parents' home to study in other city. He didn't take any cash with him, he onl ...

  4. URAL 2014 Zhenya moves from parents --线段树

    题意:儿子身无分文出去玩,只带了一张他爸的信用卡,当他自己现金不足的时候就会用信用卡支付,然后儿子还会挣钱,挣到的钱都是现金,也就是说他如果有现金就会先花现金,但是有了现金他不会还信用卡的钱.他每花一 ...

  5. 【线段树】Gym - 100507C - Zhenya moves from parents

    线段树每个结点维护两个值,分别是这个区间的 负债 和 余钱. 按时间顺序从前往后看的时候,显然负债是单调不减的. 按时间顺序从后往前看的时候,显然余钱也是单调不减的,因为之前如果有余钱,可能会增加现在 ...

  6. Gym 100507D Zhenya moves from the dormitory (模拟)

    Zhenya moves from the dormitory 题目链接: http://acm.hust.edu.cn/vjudge/contest/126546#problem/D Descrip ...

  7. ural 2015 Zhenya moves from the dormitory(模拟)

    2015. Zhenya moves from the dormitory Time limit: 1.0 secondMemory limit: 64 MB After moving from hi ...

  8. D - Zhenya moves from the dormitory URAL - 2015

    After moving from his parents’ place Zhenya has been living in the University dormitory for a month. ...

  9. NEERC 2014, Eastern subregional contest

    最近做的一场比赛,把自己负责过的题目记一下好了. Problem B URAL 2013 Neither shaken nor stirred 题意:一个有向图,每个结点一个非负值,可以转移到其他结点 ...

随机推荐

  1. crontab如何设置秒级别的定时【转载】

    * * * * * date > /home/gamester88/test/nihao.txt * * * * * (sleep 10 && date >> /ho ...

  2. centos mono

    wget http://download.mono-project.com/sources/mono/mono-4.0.1.tar.bz2wget http://download.mono-proje ...

  3. replication across two data centers

    http://andyhan.net/index.php/sys-adm/item/291-hbase-replication http://shitouer.cn/2013/04/hbase-mul ...

  4. socket小解

    要理解socket,首先得理解TCP/IP协议族, TCP/IP (Transmission Control Protocol/Internet Protocol)传输控制协议/网间协议 定义: TC ...

  5. C# tostring()汇总

    原文:http://www.cnblogs.com/xiaopin/archive/2010/11/05/1870103.html C 货币 2.5.ToString("C") ¥ ...

  6. Oulipo HDU 1686 KMP模板

    题目大意:求模式串在主串中的出现次数. 题目思路:KMP模板题 #include<iostream> #include<algorithm> #include<cstri ...

  7. 大学二三事——那些人(1)

    校歌墙的对面是一座历史比较悠久的建筑,以前叫做12号楼,后来改成了"诚"字楼. 在诚字楼一楼昏暗的走廊上,你总是能看见一位大概四五十岁的大叔,有时他会指着挂在墙上的学校简介,一个人 ...

  8. JDBC 事务隔离级别

    JDBC 事务隔离级别     先解释一下:a:脏读取:一个事务读取了另外一个并行事务未提交的数据b:不可重复读取:一个事务再次读取之前的数据时得到的数据不一致,被另外一个事务修改c:虚读:一个事务重 ...

  9. 批处理数据--db2备份数据

    如果要插入数据,前提先根据主键删除记录,然后在插入. 批处理包含两个必要文件 init.bat和start.bat 文档内容如下 init.bat内容如下 @echo ondb2 connect to ...

  10. GenericApp SampleApp SimpleAp的区别

    SampleApp3.2 Zigbee2007 协议栈实验例程表演说明C:\Texas Instruments\ZStack-2.0.0-1.2.0\Projects\zstack\Samples\S ...