POJ 2528 Mayor ' s Posters (segment tree interval update, discretization) __ Segment Tree

Topic Link: http://poj.org/problem?id=2528

Topic: On the wall to paste ads, advertising suites have a great deal of small, and advertising has successively, the back of the ads will cover the front of the ads to solve the last visible advertising surface, as shown in the following picture:




Two views, and finally the number of ads you can see from front view is 4.

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace
Std
typedef long Long LL;
	struct node{int L, R, M;
ll Val;
};
const int MAXN = 10010;
Node T[MAXN << 4];
int COL[MAXN];
int ans;  
        void pushdown (int rt) {if (T[rt].val!=-1) {t[rt << 1].val = t[rt].val;  
        T[rt << 1 | 1].val = T[rt].val;  
    T[rt].val =-1;
	} void build (int begin, int end, int rt) {T[RT].L = begin;
	T[RT].R = end;
	T[RT].M = (t[rt].l + t[rt].r) >> 1;
	T[rt].val =-1;
	if (begin = = end) return;
	Build (T[RT].L, T[RT].M, RT << 1);
Build (T[RT].M + 1, T[RT].R, RT << 1 | 1);
		} void Update (int l, int R, int num, int rt) {if (L = = T[rt].l && T[RT].R = = R) {t[rt].val = num;
	return;
	} pushdown (RT);  
    if (L > T[RT].M) Update (l, R, num, rt << 1 | 1);  
    else if (r <= t[rt].m) update (L, R, Num, RT << 1); else {update (L, T[RT].M, num, RT << 1);  
    Update (T[RT].M + 1, R, num, rt << 1 | 1);
		} void query (int rt) {if (T[rt].val!=-1) {if (!col[t[rt].val)) ans++;
		Col[t[rt].val] = 1;
	return;
	} if (t[rt].l = = T[RT].R) return;
	Query (RT << 1);
Query (RT << 1 | 1);
int x[maxn<<2], top;
Pair<int, int> Q[MAXN];
	int main () {int t;
	scanf ("%d", &t);
	int n;
	int TL, TR;
		while (t--) {top = 0;
		scanf ("%d", &n);
			for (int i = 0; i < n; i++) {scanf ("%d%d", &q[i].first, &q[i].second);
			x[top++] = Q[i].first;
		x[top++] = Q[i].second;
		Sort (x, x + top);
		int m = 1;
		for (int i = 1; i < top; i++) {if (X[i]!= x[i-1]) x[m++] = X[i];
		for (int i = m-1 i > 0; i--) {if (X[i]-1!= x[i-1]) x[m++] = 1;
		Sort (x, x + m);
		Build (0, M, 1);
			for (int i = 0; i < n; i++) {int L = lower_bound (x, X + M, Q[i].first)-X;
			int r = Lower_bound (x, X + M, q[i].second)-X;
		Update (L, R, I, 1); } memset (Col, 0, sizeof col);
		Ans = 0;
		Query (1);
	printf ("%d\n", ans);
return 0; }