POJ 2528 Mayor's posters （线段树区间更新、离散化）

题目链接：http://poj.org/problem?id=2528

题意：题目大意：在墙壁上贴广告，广告的版面有大有小，并且贴广告有先后之分，后面贴的广告会覆盖前面的广告，求解最后能看到的广告面，如下图所示：



两种视图，最后从Front View能看见的广告数目是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++] = x[i] - 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;
}