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?nn?days left before the end of the term (numbered from?11?to?nn?), and initially all of them are working days. Then the university staff sequentially publishes?qq?orders, one after another. Each order is characterised by three numbers?ll?,?rr?and?kk?:

• If?k=1k=1?, then all days from?ll?to?rr?(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=2k=2?, then all days from?ll?to?rr?(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!

輸入輸出格式

輸入格式：

?

The first line contains one integer?nn?, and the second line — one 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?qq?lines follow,?ii?-th line containing three 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?qq?integers.?ii?-th of them must be equal to the number of working days left until the end of the term after the first?iiorders are published.

?

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開點直接打標機下傳、、、畢竟線段樹一次操作涉及的節點數是log n級別的，就算是區間操作+需要標記下傳的話，
需要的空間也只是常數大了大，復雜度依然是log n的，而且很多操作還會有重復的區間呢。

于是我就動態開點+線段樹打tag瞎做了做，，，本地垃圾筆記本3s才能過極端數據，，，，不過codeforces的評測機跑的飛快，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; }

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轉載于:https://www.cnblogs.com/JYYHH/p/8406106.html

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