POJ 2528 线段树+离散化

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
#include <string>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
//#include <unordered_map>
#define Fbo friend bool operator < (node a, node b)
#define mem(a, b) memset(a, b, sizeof(a))
#define FOR(a, b, c) for(int a = b; a <= c; a++)
#define RFOR(a,b, c) for(int a = b; a >= c; a--)
#define off ios::sync_with_stdio(0)
bool check1(int a) { return (a & (a - 1)) == 0 ? true : false; }

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int Maxn = 2e5 + 5;
const double pi = acos(-1.0);
const double eps = 1e-8;

ll a[Maxn];
ll li[Maxn], ri[Maxn];
vector<ll>num;

struct node {
    ll l, r;//区间[l,r]
    ll add;//区间的延时标记
    ll sum;//区间和
}tree[Maxn * 4 + 9];
ll n;

void pushup(ll rt) { //向上更新
    tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
}
void pushdown(ll rt) {//向下更新
    if (tree[rt].add) { //当我们的标记不是0，说明该段染色，所以往下传递
        tree[rt << 1].add = tree[rt].add;
        tree[rt << 1 | 1].add = tree[rt].add; //这里不能用加
        tree[rt].add = 0;  //取消本层标记
    }
}
void BuildTree(ll l, ll r, ll rt) {
    tree[rt].l = l;
    tree[rt].r = r;
    tree[rt].add = 0;
    if (l == r) {
        //tree[rt] = {l,r,0};
        return;
    }
    ll mid = (l + r) >> 1;
    BuildTree(l, mid, rt << 1); //递归左子树
    BuildTree(mid + 1, r, (rt << 1) + 1); //递归右子树
    pushup(rt);//向上更新
}
void updata(ll l, ll r, ll val, ll rt) { //root为结点 区间更新
    //cout << l << " " << tree[rt].l << "----" << r << " " << tree[rt].r << endl;
    if (l <= tree[rt].l && r >= tree[rt].r) {
        tree[rt].add = val;//延时标记 
        return;
    }
    pushdown(rt);
    ll mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) {
        updata(l, r, val, rt << 1);
    }
    if (r > mid) {
        updata(l, r, val, rt << 1 | 1);
    }
    pushup(rt);//向上更新 可加可不加 本题并不需要根据儿子节点来更新父亲节点也能做 
}

void query(ll l, ll r, ll rt) { ////这个询问和普通线段树的询问略有不同 
    if (tree[rt].add) {   // 如果标记存在，说明这个区间是单色的 
        a[tree[rt].add] = 1;  //将有被染色的颜色记录下来 
        return;
    }
    if (tree[rt].l == tree[rt].r)  //一定要这个判断条件，否则你没有递归终止条件，就RE了 
        return;              //这与普通线段树的格式有区别，但是原理相同 
    pushdown(rt);
    ll mid = (tree[rt].l + tree[rt].r) >> 1;
    if (l <= mid) {
        query(l, r, rt << 1);//如果是0，有两种情况，一个是没有染色，一个是有多种颜色 
    }
    if (r > mid) {
        query(l, r, rt << 1 | 1);
    }
}

ll solve() { //离散化操作和插点操作 
    ll x, y;
    sort(num.begin(), num.end()); //先排序
    num.erase(unique(num.begin(), num.end()), num.end()); ////删除重复的元素
    int m = num.size();
    for (int i = 0; i < m - 1; i++) {
        if (num[i + 1] - num[i] > 1)
            num.push_back((num[i + 1] - 1));//所有相隔距离大于1的点之间都插入一个单点
    }                                       //（让它充当那段被裸露区间）,这样就不会漏掉点
    sort(num.begin(), num.end());
    BuildTree(0, num.size() - 1, 1);
    for (ll i = 1; i <= n; i++) { // 全开ll方便点
        x = lower_bound(num.begin(), num.end(), li[i]) - num.begin();
        y = lower_bound(num.begin(), num.end(), ri[i]) - num.begin();
        updata(x, y, i, 1);
    }
    ll ans = 0;
    mem(a, 0);
    query(0, num.size() - 1, 1);
    FOR(i, 1, n) {
        if (a[i])
            ans++;
    }
    return ans;
}

int main()
{
    ll t;
    scanf("%lld", &t);
    while (t--) {
        scanf("%lld", &n);
        num.clear();
        FOR(i, 1, n) {
            scanf("%lld%lld", &li[i], &ri[i]);
            num.push_back(li[i]);
            num.push_back(ri[i]);
        }
        printf("%lld\n", solve());
    }
    return 0;
}