POJ

简单翻译

n（n<=10000) 个人依次贴海报,给出每张海报所贴的范围li，ri（1<=li<=ri<=10000000) 。求出最后还能看见多少张海报。
Input
第一行: 样例个数T
第二行: 贴海报的人n
接下来n行: 每个人贴海报的范围

Output
对于每一个输入，输出最后可以看到的海报张数。下面这个图是样例解释
Sample Input
1
5
1 4
2 6
8 10
3 4
7 10
Sample Output
4

题解

显然想到线段树。因为区间太大，可以离散化，把每个点排序重新编号再放回原区间。
倒过来添加，判断，如果没有被覆盖，ans++，否则覆盖
常数巨大的丑陋代码

# include <stdio.h># include <stdlib.h># include <iostream># include <string.h># include <algorithm>using namespace std;# define IL inline# define RG register# define UN unsigned# define ll long long# define rep(i, a, b) for(RG int i = a; i <= b; i++)# define per(i, a, b) for(RG int i = b; i >= a; i--)# define uev(e, u) for(RG int e = ft[u]; e != -1; e = edge[e].nt)# define mem(a, b) memset(a, b, sizeof(a))# define max(a, b) ((a) > (b)) ? (a) : (b)# define min(a, b) ((a) < (b)) ? (a) : (b)IL ll Get(){    RG char c = '!'; RG ll num = 0, z = 1;    while(c != '-' && (c > '9' || c < '0')) c = getchar();    if(c == '-') z = -1, c = getchar();    while(c >= '0' && c <= '9') num = num*10+c-'0', c = getchar();    return num*z;}const int MAXN = 10001, MAXM = 1000001;struct Tree{    int l, r, num;} tree[MAXM];int n, ans, a[MAXN][2], num, flag;struct Post{    int v, pos;    IL bool operator <(Post c) const{        return v < c.v;    }} t[MAXN<<1];IL void Updata(RG int now){    if(!tree[now<<1].num || !tree[now<<1|1].num) return;    tree[now].num = 1;}IL void Build(RG int now, RG int l, RG int r){    tree[now] = (Tree) {l, r, 0};    if(l == r) return;    RG int mid = l+r>>1;    Build(now<<1, l, mid); Build(now<<1|1, mid+1, r);}IL void Add(RG int l, RG int r, RG int now){    if(tree[now].num) return;    if(tree[now].l == l && tree[now].r == r){        tree[now].num = 1; flag = 1;        Updata(now);        return;    }    RG int mid = tree[now].l+tree[now].r>>1;    if(mid >= r) Add(l, r, now<<1);    if(l <= mid && mid < r) Add(l, mid, now<<1), Add(mid+1, r, now<<1|1);    if(mid < l) Add(l, r, now<<1|1);    Updata(now);}int main(){    RG int T = Get();    while(T--){        n = Get(); ans = num = 0;        rep(i, 1, n){            t[i*2-1].v = Get(); t[i*2].v = Get();            t[i*2-1].pos = i*2-1;            t[i*2].pos = i*2;        }        sort(t+1, t+2*n+1);        rep(i, 1, 2*n){            RG int c = t[i].pos%2, pos = (t[i].pos+1)/2;            if(t[i].v != t[i-1].v) num++;            a[pos][!c] = num;        }        Build(1, 1, num);        per(i, 1, n){            flag = 0;            Add(a[i][0], a[i][1], 1);            ans += flag;        }        printf("%d\n", ans);    }    return 0;}