Billionaires
Time limit: 3.0 second
Memory limit: 64 MB
Memory limit: 64 MB
You probably are aware that Moscow holds the first place in the world with respect to the number of billionaires living there. However, the work of billionaires is such that they have to travel a lot. That is why some other city can be the first in such a list on certain days. Your friends from FSB, FBI, MI5, and Shin Bet have provided you with information about movements of billionaires during some period of time. Your employer asks you to determine for each city the number of days during this period on which this city exceeded all other cities in the total amount of money that billionaires staying in this city have.
Input
In the first line you are given the number n of billionaires (1 ≤ n ≤ 10000). The following nlines contain information about these people: their names, cities where they were staying at the beginning of the period, and their fortunes. In the next line you are given the number m of days in the period for which you have the information (1 ≤ m ≤ 50000) and the number k of travels of the billionaires (0 ≤ k ≤ 50000). The following k lines contain the list of travels in the following format: the number of the day (from 1 to m−1), the name of the person, and the city of destination. You may assume that billionaires depart late at night and arrive to the destination city on the next day's morning. They cannot make more than one travel each day. The numbers of days in the list are not decreasing. All names of people and cities consist of at most 20 English letters; you must take into consideration the case of the symbols. The fortunes are in the range from 1 to 100 billions (one billion is a thousand million).
Output
In each line of the output give the name of a city and, after a space, the number of days during which this city was the first with respect to the sum of fortunes of the billionaires staying there. Leave out those cities for which there were no such days. The cities must be sorted alphabetically (with the usual symbol order: ABC...Zabc...z).
Sample
| input | output |
|---|---|
5 |
Anadyr 5 |
分析：线段树单点更新查询最大值坐标；
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+;
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;p=p*p;q>>=;}return f;}
int n,m,k,t,d,now[maxn],num,cnt;
map<string,int>ci;
map<string,int>id;
map<int,string>to;
ll mo[maxn];
struct node
{
string x,y;
}a[maxn];
struct node1
{
int t;
string x,y;
}q[maxn];
struct node2
{
int id,t;
bool operator<(const node2&p)const
{
return to[id]<to[p.id];
}
}ans[maxn];
struct Node
{
ll Max, lazy;
} T[maxn<<]; void PushUp(int rt)
{
T[rt].Max = max(T[rt<<].Max, T[rt<<|].Max);
} void PushDown(int L, int R, int rt)
{
int mid = (L + R) >> ;
ll t = T[rt].lazy;
T[rt<<].Max += t;
T[rt<<|].Max += t;
T[rt<<].lazy += t;
T[rt<<|].lazy += t;
T[rt].lazy = ;
} void Update(int l, int r, ll v, int L, int R, int rt)
{
if(l==L && r==R)
{
T[rt].lazy += v;
T[rt].Max += 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);
} int Query(int L, int R, int rt)
{
if(L==R)return L;
if(T[rt].lazy)PushDown(L,R,rt);
int mid=L+R>>;
if(T[rt<<].Max>T[rt<<|].Max)return Query(Lson);
else if(T[rt<<].Max<T[rt<<|].Max)return Query(Rson);
else return ;
}
int main()
{
int i,j;
scanf("%d",&n);
rep(i,,n)
{
cin>>a[i].x>>a[i].y>>mo[i];
id[a[i].x]=i;
if(!ci[a[i].y])ci[a[i].y]=++num,to[num]=a[i].y;
now[i]=ci[a[i].y];
}
scanf("%d%d",&d,&m);
rep(i,,m)
{
cin>>q[i].t>>q[i].x>>q[i].y;
if(!ci[q[i].y])ci[q[i].y]=++num,to[num]=q[i].y;
}
rep(i,,num)ans[i].id=i;
rep(i,,n)Update(ci[a[i].y],ci[a[i].y],mo[i],,num,);
j=;
rep(i,,d)
{
while(j<=m&&q[j].t==i-)
{
Update(now[id[q[j].x]],now[id[q[j].x]],-mo[id[q[j].x]],,num,);
Update(ci[q[j].y],ci[q[j].y],mo[id[q[j].x]],,num,);
now[id[q[j].x]]=ci[q[j].y];
j++;
}
if((cnt=Query(,num,)))ans[cnt].t++;
}
sort(ans+,ans+num+);
rep(i,,num)if(ans[i].t)printf("%s %d\n",to[ans[i].id].c_str(),ans[i].t);
//system("Pause");
return ;
}