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Codeforces 915 E Physical Education Lessons

發(fā)布時間：2023/11/30 编程问答 37 豆豆

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題目描述

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?

Help Alex to determine the number of working days left after each order!

輸入輸出格式

輸入格式：

The first line contains one integer? $integer?qq?(?1<=n<=10^{9}1<=n<=109?,?1<=q<=3·10^{5}1<=q<=3?105) — the number of days left before the end of the term, and the number of orders, respectively.$

Then? $integers?l_{i}li??,?r_{i}ri??and?k_{i}ki??representing?ii?-th order (?1<=l_{i}<=r_{i}<=n1<=li?<=ri?<=n?,?1<=k_{i}<=21<=ki?<=2?).$

輸出格式：

Print?

輸入輸出樣例

輸入樣例#1：?
4 6 1 2 1 3 4 1 2 3 2 1 3 2 2 4 1 1 4 2 輸出樣例#1：? 2 0 2 3 1 4

下午去湘江邊的橘子洲van了一圈，然而明天就要去雅禮報道了hhhh

很明顯n<=10^9肯定不能用普通的線段樹做，雖然可能離散化完了之后會有奇淫技巧能做。。。。
但是我首先想到的是動態(tài)開點直接打標(biāo)機(jī)下傳、、、畢竟線段樹一次操作涉及的節(jié)點數(shù)是log n級別的，就算是區(qū)間操作+需要標(biāo)記下傳的話，
需要的空間也只是常數(shù)大了大，復(fù)雜度依然是log n的，而且很多操作還會有重復(fù)的區(qū)間呢。

于是我就動態(tài)開點+線段樹打tag瞎做了做，，，本地垃圾筆記本3s才能過極端數(shù)據(jù)，，，，不過codeforces的評測機(jī)跑的飛快，硬生生的縮成了300+ms。

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cmath> #define ll long long #define maxn 200005 using namespace std; struct node{int lc,rc;int tag,sum; }nil[maxn*71]; int n,le,ri,tot; int q,opt,cnt=1;inline void work(int v,int tmp,int len){if(tmp==-1) nil[v].tag=-1,nil[v].sum=0;else nil[v].tag=1,nil[v].sum=len; }inline void pushdown(int o,int l,int r){int ls=nil[o].lc,rs=nil[o].rc,mid=l+r>>1;if(nil[o].tag==-1){nil[o].tag=0;work(ls,-1,mid-l+1);work(rs,-1,r-mid);}else if(nil[o].tag==1){nil[o].tag=0;work(ls,1,mid-l+1);work(rs,1,r-mid);} }void update(int u,int l,int r){if(l>=le&&r<=ri){int pre=nil[u].sum;work(u,opt,r-l+1);tot+=nil[u].sum-pre;return;}if(!nil[u].lc) nil[u].lc=++cnt;if(!nil[u].rc) nil[u].rc=++cnt;pushdown(u,l,r);int mid=l+r>>1;if(le<=mid) update(nil[u].lc,l,mid);if(ri>mid) update(nil[u].rc,mid+1,r);nil[u].sum=nil[nil[u].lc].sum+nil[nil[u].rc].sum; }int main(){ // freopen("data.in","r",stdin); // freopen("data.out","w",stdout);int root=1;nil[1]=(node){0,0,0,0};scanf("%d%d",&n,&q);while(q--){scanf("%d%d%d",&le,&ri,&opt);opt=(opt==2?-1:1);update(root,1,n);printf("%d\n",n-tot);}return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/JYYHH/p/8406106.html

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