Physical Education Lessons CodeForces - 915E （动态开点线段树）

Physical Education Lessons

CodeForces - 915E

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!

Input

The first line contains one integer n, and the second line — one integer q (1 ≤ n ≤ 109, 1 ≤ q ≤ 3·105) — the number of days left before the end of the term, and the number of orders, respectively.

Then q lines follow, i-th line containing three integers l**i, r**i and k**i representing i-th order (1 ≤ l**i ≤ r**i ≤ n, 1 ≤ k**i ≤ 2).

Output

Print q integers. i-th of them must be equal to the number of working days left until the end of the term after the first i orders are published.

Example

Input

461 2 13 4 12 3 21 3 22 4 11 4 2

Output

202314

思路：

区间很大，询问不是很大，经典的动态开点线段树问题，

即在处理询问的过程中，需要用到一个区间再到线段树中去申请一个位置给它。

大概需要开不到\(2*q*log(x)\) 左右个节点即可完成。x是区间的最大长度。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}

inline void getInt(int *p);
const int maxn = 300010 * 30 * 2;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
struct node {
    int lson;
    int rson;
    int val;
    int laze;
} segment_tree[15001000 ];
int n, m;
int tot = 1;
void pushdown(int rt, int l, int r, int mid)
{
    if (segment_tree[rt].val) {
        segment_tree[segment_tree[rt].lson].val = (mid - l + 1);
        segment_tree[segment_tree[rt].rson].val = ( r - mid  );
    } else {
        segment_tree[segment_tree[rt].lson].val = 0;
        segment_tree[segment_tree[rt].rson].val = 0;
    }
    segment_tree[rt].laze = 0;
    segment_tree[segment_tree[rt].lson].laze = 1;
    segment_tree[segment_tree[rt].rson].laze = 1;
}
void update(int rt, int ql, int qr, int l, int r, int op)
{
    if (l <= ql && qr <= r) {
        segment_tree[rt].laze = 1;
        segment_tree[rt].val = (qr - ql + 1) * op;
        return;
    }
    if (!segment_tree[rt].lson) {
        segment_tree[rt].lson = ++tot;
    }
    if (!segment_tree[rt].rson) {
        segment_tree[rt].rson = ++tot;
    }
    if (segment_tree[rt].laze) {
        pushdown(rt, ql, qr, qr + ql >> 1);
    }
    int mid = (ql + qr) >> 1;
    if (l <= mid) {
        update(segment_tree[rt].lson, ql, mid, l, r, op);
    }
    if (r > mid) {
        update(segment_tree[rt].rson, mid + 1, qr, l, r, op);
    }
    int ls = segment_tree[rt].lson;
    int rs = segment_tree[rt].rson;
    segment_tree[rt].val = segment_tree[ls].val + segment_tree[rs].val;
}
int main()
{
    //freopen("D:\\code\\text\\input.txt","r",stdin);
    //freopen("D:\\code\\text\\output.txt","w",stdout);
    du2(n, m);
    while (m--) {
        int l, r, op;
        du3(l, r, op);
        if (op == 1) {
            update(1, 1, n, l, r, op);
        } else {
            update(1, 1, n, l, r, 0);
        }
        printf("%d\n", n - segment_tree[1].val );
    }
    return 0;
}

inline void getInt(int *p)
{
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    } else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}