zoukankan      html  css  js  c++  java
  • Educational Codeforces Round 36 (Rated for Div. 2) E. Physical Education Lessons(动态开点线段树)

    链接:

    https://codeforces.com/problemset/problem/915/E

    题意:

    This year Alex has finished school, and now he is a first-year student of Berland State University. For him it was a total surprise that even though he studies programming, he still has to attend physical education lessons. The end of the term is very soon, but, unfortunately, Alex still hasn't attended a single lesson!

    Since Alex doesn't want to get expelled, he wants to know the number of working days left until the end of the term, so he can attend physical education lessons during these days. But in BSU calculating the number of working days is a complicated matter:

    There are n days left before the end of the term (numbered from 1 to n), and initially all of them are working days. Then the university staff sequentially publishes q orders, one after another. Each order is characterised by three numbers l, r and k:

    If k = 1, then all days from l to r (inclusive) become non-working days. If some of these days are made working days by some previous order, then these days still become non-working days;
    If k = 2, then all days from l to r (inclusive) become working days. If some of these days are made non-working days by some previous order, then these days still become working days.
    Help Alex to determine the number of working days left after each order!

    思路:

    求1-n里的0的个数。
    动态开点线段树,对于每次更新,每个节点先判断是不是新的,是新的就重新建立即可。

    代码:

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    const int MAXN = 4e5+10;
    const int MOD = 1e9+7;
    const int INF = 1e9;
    
    int n, q;
    int add[MAXN*40], sum[MAXN*40], ln[MAXN*40], rn[MAXN*40];
    int rot = 0, cnt = 0;
    
    void pushdown(int root, int l, int r)
    {
        if (add[root] != -1)
        {
            if (!ln[root]) ln[root] = ++cnt;
            if (!rn[root]) rn[root] = ++cnt;
            int mid = (l+r)/2;
            add[ln[root]] = add[root];
            add[rn[root]] = add[root];
            sum[ln[root]] = add[root]*(mid-l+1);
            sum[rn[root]] = add[root]*(r-mid);
            add[root] = -1;
        }
    }
    
    void update(int &root, int l, int r, int ql, int qr, int v)
    {
        if (!root) root = ++cnt;
        if (ql <= l && r <= qr)
        {
            add[root] = v;
            sum[root] = v*(r-l+1);
            return;
        }
        pushdown(root, l, r);
        int mid = (l+r)/2;
        if (ql <= mid) update(ln[root], l, mid, ql, qr, v);
        if (qr > mid) update(rn[root], mid+1, r, ql, qr, v);
        sum[root] = sum[ln[root]]+sum[rn[root]];
    }
    
    int main()
    {
        //dsu on tree 启发式合并
        ios::sync_with_stdio(stdin);
        cin.tie(0), cout.tie(0);
        scanf("%d%d", &n, &q);
        memset(add, -1, sizeof(add));
        int l, r, k;
        while(q--)
        {
            scanf("%d%d%d", &l, &r, &k);
            update(rot, 1, INF, l, r, k == 2 ? 0 : 1);
            printf("%d
    ", n-sum[1]);
        }
    
        return 0;
    }
    
  • 相关阅读:
    Objective-C NSFileManager 文件管理总结
    ScrollView在RelativeLayout失效问题
    解决myeclipse中struts2 bug问题包的替换问题
    SOA究竟是个啥
    Flash--元件和实例
    MyEclipse中加入web项目到tomcat
    [C]if (CONDITION)语句中CONDITION的情况
    GTK经常使用控件之笔记本控件( GtkNotebook )
    org.apache.solr.handler.dataimport.DataImportHandlerException: Data Config problem: 对实体 &quot;characterEn
    Android自动测试之Monkey工具
  • 原文地址:https://www.cnblogs.com/YDDDD/p/12238018.html
Copyright © 2011-2022 走看看