Kattis - gcpc (treap模板)

ne hundred years from now, in 21172117, 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 nn teams in the contest. The teams are numbered 1,2,…,n1,2,…,n, and your favorite team has number 11.

Like today, the score of a team is a pair of integers (a,b)(a,b)where aa is the number of solved problems and bb 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 t1t1 and t2t2 whose scores are (a1,b1)(a1,b1) and (a2,b2)(a2,b2). The score of team t1t1 is better than that of t2t2if either a1>a2a1>a2, or if a1=a2a1=a2 and b1<b2b1<b2. The rank of a team is k+1k+1 where kk 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.

Input

The first line of input contains two integers nn and mm, where 1≤n≤1051≤n≤105 is the number of teams, and 1≤m≤1051≤m≤105 is the number of events.

Then follow mm lines that describe the events. Each line contains two integers tt and pp (1≤t≤n1≤t≤n and 1≤p≤10001≤p≤1000), meaning that team tt has solved a problem with penalty pp. The events are ordered by the time when they happen.

Output

Output mm lines. On the ii’th line, output the rank of your favorite team after the first ii events have happened.

Sample Input 1

3 4
2 7
3 5
1 6
1 9

Sample Output 1

2
3
2
1

题意:好多队伍比赛,按照事件发生顺序给你队伍AC题目的信息,x队AC一道题,罚时为y,每有一个队过题输出编号为1的队伍的排名
我们每当每个队A题的时候把他原来状态删除,把新状态加入到treap.由于可能有好几个队A题数罚时都一样,这样按照队伍编号升序
我们每个节点上我们设置有多个队伍就行了

 #include <bits/stdc++.h>
 #define rep(i,a,n) for (int i=a;i<=n;i++)
 #define per(i,a,n) for (int i=n;i>=a;i--)
 #define pb push_back
 #define mp make_pair
 #define all(x) (x).begin(),(x).end()
 #define fi first
 #define se second
 #define SZ(x) ((int)(x).size())
 using namespace std;
 typedef vector<int> vi;
 typedef long long ll;
 typedef pair<int,int> pii;
 const ll mod=;
 ll powmod(ll a,ll b) {ll res=;a%=mod; assert(b>=); for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
 const int maxn=1e5+;
 struct Node
 {
     Node *ch[];//0 左儿子(名次靠前) 1右儿子(名次靠后)
     int r;//随机优先级,数值越大优先级越高
     int v;//值
     int s;//以他为根的子树上队伍数的个数
     int num;//当前节点上队伍数目(有队伍会题数罚时都相同)一个节点可以右多个队伍
     int tim;//罚时
     Node(int v,int tim):v(v),tim(tim){ch[]=ch[]=NULL;r=((rand()<<)|rand());s=;num=; }
     inline int cmp(int x,int xtim) const {//比较函数 相等-1 名次靠前0(左子树) 名次靠后 1(右子树)
         if(x==v&&tim==xtim) return -;//相等
         if(x==v) return xtim<tim?:;//题数相同罚时少的优先
         return x>v?:;
     }
     inline void maintain() {
         s=num;//当前节点子树个数
         if(ch[]!=NULL) s+=ch[]->s;//加上自己左子树
         if(ch[]!=NULL) s+=ch[]->s;//加上自己右子树
     }
 };
 void  rotate(Node* &o,int d) {//旋转
     Node* k=o->ch[d^];o->ch[d^]=k->ch[d];k->ch[d]=o;
     o->maintain();
     k->maintain();
     o=k;
 }
 void insert(Node* &o,int x,int tim) {//将 过了x题罚时tim的队伍插入到treap
     if(o==NULL) {
         o=new Node(x,tim);
     }
     else {
         int d;
         if(x>o->v) {//题数多 查到左子树
             d=;
         }
         else if(x==o->v) {//题数一样
             if(tim<o->tim) d=;//罚时少 左子树
             else d=;//罚时多 右子树
         }
         else d=;//题数少 右子树
         if(x==o->v&&tim==o->tim) {//找到当前自己题数罚时都一样的节点
             (o->s)++;//当前节点子树的节点上的队伍数++
             (o->num)++;//在该节点的队伍数++
         }
         else  {
             insert(o->ch[d],x,tim);if(o->ch[d]->r > o->r) rotate(o,d^);
         }
     }
     o->maintain();
 }
 void remove(Node* &o,int x,int tim) {
     int d=o->cmp(x,tim);
     if(d==-) {
         if(o->num==) {//如果处于当前题数罚时的队伍就一个队
             Node* u=o;
             if(o->ch[]!=NULL&&o->ch[]!=NULL) {
                 int d2=(o->ch[]->r>o->ch[]->r?:);
                 rotate(o,d2);
                 remove(o->ch[d2],x,tim);
             }
             else {
                 if(o->ch[]==NULL) o=o->ch[];
                 else o=o->ch[];
                 delete u;
             }
         }
         else {//否则直接当前节点队伍数--
             (o->num)--;
         }
     }
     else {
         remove(o->ch[d],x,tim);
     }
     if(o!=NULL) o->maintain();
 }
 int Rank(Node* o,int x,int tim) {
     if(o==NULL) return ;
     int s=(o->ch[]==NULL?:o->ch[]->s);//s是o的左子树上队伍的个数
     if(o->v==x&&o->tim==tim) {//找到当前队伍状态的节点
         return s+;//返回左子树队伍数+1
     }
     else if(o->v<x||(o->v==x&&o->tim>tim)) {//排名在当前节点之前,往左子树找
         return Rank(o->ch[],x,tim);
     }
     else {//排名在当前节点之后,排名为当前节点左子树队伍数+当前节点队伍数+往右子树找
         return Rank(o->ch[],x,tim)+s+(o->num);
     }
 }
 int nownum[maxn],nowtim[maxn];
 int main()
 {
     Node *root=NULL;//初始化
     int n,m;
     scanf("%d%d",&n,&m);
     int u,v;
     rep(i,,n) insert(root,,);//大家一开始都没过题
     rep(i,,m) {
         scanf("%d%d",&u,&v);
         remove(root,nownum[u],nowtim[u]);//移除当前队伍之前的状态
         nownum[u]++;//题数加1
         nowtim[u]+=v;//加罚时
         insert(root,nownum[u],nowtim[u]);//插入当前状态
         printf("%d\n",Rank(root,nownum[],nowtim[]));//查询队伍编号为1的名次
     }
     return ;
 }
 /**
 3 10
 2 1
 2 1
 3 1
 1 1
 3 5
 3 5
 1 1
 1 2
 3 1

 */