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

来源：互联网发布：福州数据分析培训编辑：程序博客网时间：2024/05/01 16:00

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;}

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