[CF# 366 Thor] STL模拟

[CF# 366 Thor] STL模拟

题目链接： [CF# 366 Thor] 

题意描述：N 种物品标号为1~N， Q次操作：

1. 标号为x 的物品增加一个；
2. 将标号为x的物品全部删除；
3. 将最初加入的前t次物品全部删除。

解题思路：想了好久，就是想不到模拟。 竟然一直想着用树状数组求和....

直接看官方题解吧....

Consider a queue e for every application and also a queue Q for the notification bar. When an event of the first type happens, increase the number of unread notifications by 1 and push pair (i, x) to Q where i is the index of this event among events of the first type, and also push number i to queue e[x].

When a second type event happens, mark all numbers in queue e[x] and clear this queue (also decreese the number of unread notifications by the number of elements in this queue before clearing).

When a third type query happens, do the following:

while Q is not empty and Q.front().first <= t:
	i = Q.front().first
	x = Q.front().second
	Q.pop()
	if mark[i] is false:
		mark[i] = true
		e[v].pop()
		ans = ans - 1 // it always contains the number of unread notifications

But in C++ set works much faster than queue!

Time Complexity: 

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define fst             first
#define snd             second
#define lson            l, mid, rt << 1
#define rson            mid + 1, r, rt << 1 | 1

typedef __int64  LL;
//typedef long long LL;
typedef unsigned int uint;
typedef pair PII;

const int MAXN = 300000 + 5;

int N, Q_CNT;
int type, res;

priority_queue, greater > Q;
queue E[MAXN];
bool vis[MAXN];

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif // ONLINE_JUDGE
    int x, t, id;
    PII top;
    while (~scanf ("%d %d", &N, &Q_CNT) ) {
        res = id = 0;
        while (!Q.empty() ) Q.pop();
        for (int i = 1; i <= N; i++) while (!E[i].empty() ) E[i].pop();
        memset (vis, false, sizeof (vis) );
        for (int i = 1; i <= Q_CNT; i ++) {
            scanf ("%d", &type);
            if (type == 1) {
                scanf ("%d", &x);
                ++ id;
                Q.push (PII (id, x) );
                E[x].push (id);
                ++ res;
            } else if (type == 2) {
                scanf ("%d", &x);
                while (!E[x].empty() ) {
                    t = E[x].front();
                    E[x].pop();
                    if (!vis[t]) {
                        vis[t] = true;
                        -- res;
                    }
                }
            } else {
                scanf ("%d", &t);
                while (!Q.empty() ) {
                    top = Q.top();
                    if (top.fst > t) break;
                    if (!vis[top.fst]) {
                        vis[top.fst] = true;
                        -- res;
                    }
                    Q.pop();
                }
            }
            printf ("%d\n", res);
        }
    }
    return 0;
}