POJ 2528 Mayor's posters（离散+线段树）

题目大意：往一面墙上贴与墙等高的海报，n次贴完后，求可以看见的海报总数（看见一部分也算）

思路：明显的区间维护，用线段树，不过裸的线段树超时超空间，可以把坐标离散，得到不超过200000个有效点，每个点都表示一个小区间（a[i]~a[i+1]这一段），然后就可以轻松地解决了。不过题目有个坑，给定的右坐标其实不是实际坐标，还要+1，区间总数等于有效点数 -1,更新的时候把查找到的右端点-1再代入，具体见代码：

/**********************************************
***    Problem:
***    Author:        JKL
***    University:    CSUST
***    Team:          __Dream
***    Email:          1451108308@QQ.COM
***    My Blog:        http://www.cnblogs.com/jklongint/
***********************************************/
//===================================================
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <numeric>
#include <ctime>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <set>
#include <bitset>
#include <deque>
using namespace std;
//---------------------------------------------------
#define mem(a,b) memset(a,b,sizeof(a))
#define GO cout<<"HelloWorld!"<<endl
#define Case(x) cout<<"Case "<<x<<":"
#define foru(i,n) for(int i=1; i <= n; i++)
#define ford(i,n) for(int i = n; i >= 1; i--)
 #define fin freopen("input.txt","r",stdin);
 #define fout freopen("output.txt","w",stdout)
#define lson  l, m, rt << 1
#define rson  m + 1, r, rt << 1 | 1

#define sqr(a)  ((a)*(a))
#define abs(a) ((a>0)?(a):-(a))
#define pii pair<int,int>

#define fmax(a,b) max(a,b)
#define fmin(a,b) min(a,b)
#define fmax3(a,b,c)  (fmax(a,fmax(a,b)))
#define fmin3(a,b,c)  (fmin(a,fmin(a,b)))

#define sfi(x) scanf("%d",&x)
#define sfL(x) scanf("%I64d",&x)
#define sfc(x) scanf("%c",&x)
#define sfd(x) scanf("%lf",&x)
#define sfs(x) scanf("%s",x)
#define sfii(a,b) scanf("%d%d",&a,&b)
#define sfLL(a,b) scanf("%I64d%I64d",&a,&b)
#define sfcc(a,b) scanf("%c%c",&a,&b)
#define sfdd(a,b) scanf("%lf%lf",&a,&b)
#define sfss(a,b) scanf("%s%s",a,b)

#define pfi(x) printf("%d",x)
#define pfL(x) printf("%I64d",x)
#define pfs(x) printf("%s",x)
#define pfd(x) printf("%lf",x)
#define pfc(x) print("%c",x)
#define newLine pfs("
")
#define space pfs(" ")

//--------------------------------------------------------
typedef long long LL;
typedef unsigned long long ULL;
typedef __int64 __LL;
typedef unsigned __int64 __ULL;

typedef vector<int> vi;
typedef vector<LL> vL;
typedef vector<string> vs;
typedef set<int> si;
typedef map<int,int> mii;
typedef map<LL,LL> mLL;
typedef map<string,int> msi;
typedef map<char,int> mci;
//--------------------------------------------------------
const int dx[4]={1,-1,0,0};
const int dy[4]={0,0,1,-1};
 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 const int N6=1000006;
 const int N5=100006;
 const int N4=10006;
 const int N3=1006;
 const int N2=106;
 const int N=20009;
 const int MOD=1000000007;
 const LL LMAX=0x7fffffffffffffff;
 const LL IMAX=0x3fffffff;
 const double PI=3.14159265359;
//--------------------------------------------------------
template< class T > T gcd(T a, T b) { return (b != 0 ? gcd<T>(b, a%b) : a); }
template< class T > T lcm(T a, T b) { return (a / gcd<T>(a, b) * b); }

//------------------------------------------------------------
struct TreeNode{
    int  mark;
};
//=================================================================
TreeNode tree[N << 2];
int T, n, cnt, flag[N], x[N], y[N], a[N], b[N];
void PushUP(int rt)
{
    int tl = tree[rt << 1].mark, tr = tree[rt << 1 | 1].mark;
    if(tl != tr || tl == -1) tree[rt].mark = -1;
    else tree[rt].mark = tl;
}
void PushDOWN(int rt)
{
    int rl = rt << 1, rr = rt << 1 | 1;
    if(tree[rt].mark > 0){
        tree[rl].mark = tree[rt].mark;
        tree[rr].mark = tree[rt].mark;
    }
}
void build(int l, int r, int rt)
{
    tree[rt].mark = 0;
    if(l == r)return;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
}
void update(int L, int R, int p, int l, int r, int rt)
{
    if(L <= l && R >= r){
        tree[rt].mark = p;
        return;
    }
    int m = (l + r) >> 1;
    PushDOWN(rt);
    if(L <= m) update(L, R, p, lson);
    if(R > m) update(L, R, p, rson);
    PushUP(rt);
}
int query(int p, int l, int r, int rt)
{
    if(l == r){
        return tree[rt].mark;
    }
    int m = (l + r) >> 1;
    PushDOWN(rt);
    if(p <= m)return query(p, lson);
    else return query(p, rson);
}
void discrete()
{
    cnt = 0;
    foru(i, n){
        a[++cnt] = x[i];
        a[++cnt] = y[i];
    }
    sort(a + 1, a + 1 + cnt);
    int cc = 0;
    foru(i, cnt){
        if(a[i] != a[i - 1] || i == 1)b[++cc] = a[i];
    }
    cnt = cc;
}
int get(int x)
{
    int l = 1, r = cnt;
    while(l < r){
        int m = (l + r) >> 1;
        if(x < b[m]) r = m - 1;
        if(x == b[m]) return m;
        if(x > b[m]) l = m + 1;
    }
    return l;
}
int main()
{
    //fin;//fout;//freopen("input.txt","r",stdin);
   cin>>T;
   while(T--){
        cin>>n;
        foru(i, n)sfii(x[i], y[i]),y[i]++;
        discrete();
        build(1, cnt - 1, 1);
        foru(i, n){
            int l = get(x[i]), r = get(y[i]) - 1;
            if(r >= l)update(l, r, i, 1, cnt - 1, 1);
        }
        mem(flag, 0);
        int ans = 0;
        foru(i, cnt - 1){
            int c = query(i, 1, cnt - 1, 1);
            if(!flag[c] && c > 0){
                flag[c] = 1;
                ans++;
            }
        }
        cout<<ans<<endl;
   }
    return 0;
}