ural2014 Zhenya moves from parents

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+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,c[maxn];
struct Node
{
    ll Min, lazy;
} T[maxn<<2];
void PushUp(int rt)
{
    T[rt].Min = min(T[rt<<1].Min, T[rt<<1|1].Min);
}
void PushDown(int L, int R, int rt)
{
    int mid = (L + R) >> 1;
    ll t = T[rt].lazy;
    T[rt<<1].Min += t;
    T[rt<<1|1].Min +=  t;
    T[rt<<1].lazy += t;
    T[rt<<1|1].lazy += t;
    T[rt].lazy = 0;
}
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) >> 1;
    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+1, 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) >> 1;
    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 + 1, r, Rson));
}
struct node
{
    int x,y;
}a[maxn];
int main()
{
    int i,j;
    scanf("%d",&n);
    rep(i,1,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*31)*24*60+d*60+e;
        c[i]=a[i].y;
    }
    sort(c+1,c+n+1);
    rep(i,1,n)a[i].y=lower_bound(c+1,c+n+1,a[i].y)-c;
    rep(i,1,n)
    {
        Update(a[i].y,n,a[i].x,1,n,1);
        printf("%lld
",min(0LL,Query(1,n,1,n,1)));
    }
    //system("pause");
    return 0;
}