7635: Galactic Collegiate Programming Contest
时间限制: 2 Sec 内存限制: 128 MB
提交: 157 解决: 51
[提交] [状态] [讨论版] [命题人:admin]
题目描述
One hundred years from now, in 2117, the International Collegiate Programming Contest (of which the NCPC is a part) has expanded significantly and it is now the Galactic Collegiate Programming Contest (GCPC).
This year there are n teams in the contest. The teams are numbered 1, 2, . . . , n, and your favorite team has number 1.
Like today, the score of a team is a pair of integers (a, b) where a is the number of solved problems and b is the total penalty of that team.
When a team solves a problem there is some associated penalty (not necessarily calculated in the same way as in the NCPC – the precise details are not important in this problem). The total penalty of a team is the sum of the penalties for the solved problems of the team.
Consider two teams t 1 and t 2 whose scores are (a1 , b1 ) and (a2 , b2 ). The score of team t 1 is better than that of t 2 if either a1 > a2 , or if a1 = a2 and b1 < b2 . The rank of a team is k + 1 where k is the number of teams whose score is better.
You would like to follow the performance of your favorite team. Unfortunately, the organizers of GCPC do not provide a scoreboard. Instead, they send a message immediately whenever a team solves a problem.
输入
The first line of input contains two integers n and m, where 1 ≤ n ≤ 105 is the number of teams,and 1 ≤ m ≤ 105 is the number of events.
Then follow m lines that describe the events. Each line contains two integers t and p (1 ≤ t ≤ n and 1 ≤ p ≤ 1000), meaning that team t has solved a problem with penalty p. The events are ordered by the time when they happen.
输出
Output m lines. On the i’th line, output the rank of your favorite team after the first i events have happened.
样例输入
3 4
2 7
3 5
1 6
1 9样例输出
2
3
2
1来源/分类
#include<bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define rep(i,j,k) for(int i=j;i<k;i++)
const int maxn = 100005;
typedef long long ll;
typedef struct
{
int num;
int pent;
}Team;
Team team[maxn];
int vis[maxn];
queue <int> q;
int main(void)
{
IO
memset(team,0,sizeof(team));
memset(vis,0,sizeof(vis));
int n,m;
scanf("%d%d",&n,&m);
int t,p;
int rank=1;
while(m--)
{
scanf("%d%d",&t,&p);
team[t].num++;
team[t].pent+=p;
if(t!=1) { //不是1队的只管比1大的情况
if(!vis[t])
if(team[t].num>team[1].num||(team[t].num==team[1].num&&team[t].pent<team[1].pent))
{
rank++;
vis[t]=1;
q.push(t);
}
printf("%d\n",rank);
continue;
}
else{ //1队A题后重新筛选队列里比1队排名要高的队伍
int len = q.size(); //必须要在循环外面确定长，不然死循环。
for(int i=0;i<len;i++)
{
int k=q.front();
q.pop();
if(team[k].num>team[1].num||(team[k].num==team[1].num&&team[k].pent<team[1].pent))
q.push(k);
else
vis[k]=0;
}
}
rank = q.size()+1;
printf("%d\n",rank);
}
}