POJ 2528 Mayor's posters [ 离散化 + 线段树 ] Mayor's posters

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我是靠谱客的博主矮小月光，最近开发中收集的这篇文章主要介绍POJ 2528 Mayor's posters [ 离散化 + 线段树 ] Mayor's posters，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

题目链接：

Mayor's posters POJ 2528

题意概括：

依次贴上 n 张海报，每张海报会覆盖一个区间。后贴上的海报会覆盖前贴的海报，问最后可以看见几张海报

这里的区间不是拿两端点来维护的，是直接按单位最小区间来编号。如 [3, 5] 区间是由编号为 3、4、5的区间组成的

数据范围：

$1leq nleq 10000$

$1leq ileq n$

$1leq l_{i}leq r_{i} leq 10000000$

题解分析：

其实就是一个区间涂色问题，能看见几张海报问的就是有几种颜色，数据结构用线段树

线段树维护的值是区间的颜色：

如果区间内的所有线段颜色相同，则该区间的颜色就是这个颜色
反之，该区间的颜色编号就是 INF，表示区间颜色不一致

之前写过的类似的题：https://blog.csdn.net/Originum/article/details/82946613

但是，不能直接建树。因为区间的长度上限达 10000000，就算是 O(nlogn) 也会TLE。还可能MLE

由于海报的数量上限为 1e4 ，我们可以用离散化方法预处理数据，把区间的长度压缩压缩到 1e4

关于离散化的方法：https://blog.csdn.net/Originum/article/details/82950423

注意的点：

这题的数据有问题，n 明显大于 10000，我之前一直RE，把数组开到 4e4 才过的
当 n == 0 时要特判
处理好离散化的bug，用加点的方法

AC代码：

#include <stdio.h>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAXN = 4e4 + 10;
const int INF = 0x3f3f3f3f;
int origin[MAXN], tree[MAXN<<2], lazy[MAXN<<2], ans[MAXN], last;
int disc[MAXN], L[MAXN], R[MAXN];
//离散数组，左端点，右端点
void pushup(int p) {
if (tree[p << 1] == tree[p << 1 | 1])
tree[p] = tree[p << 1];
else
tree[p] = INF;
}
void pushdown(int p, int l, int r) {
if (lazy[p] != INF) {
lazy[p << 1] = lazy[p];
lazy[p << 1 | 1] = lazy[p];
tree[p << 1] = lazy[p];
//左右儿子结点均按照需要加的值总和更新结点信息
tree[p << 1 | 1] = lazy[p];
lazy[p] = INF;
}
}
void build(int p, int l, int r) {
if (l == r) {
tree[p] = origin[l];
return;
}
int mid = (l + r) >> 1;
build(p << 1, l, mid);
build(p << 1 | 1, mid + 1, r);
pushup(p);
}
void update(int p, int l, int r, int ql, int qr, int v) {
//区间更新
if (ql <= l && r <= qr) {
//当前区间在更新区间内
tree[p] = v;
lazy[p] = v;
return;
}
int mid = (l + r) >> 1;
pushdown(p, l, r);
if (ql <= mid)
update(p << 1, l, mid, ql, qr, v);
if (qr > mid)
update(p << 1 | 1, mid + 1, r, ql, qr, v);
pushup(p);
}
void query(int p, int l, int r, int ql, int qr) {
if (tree[p] != INF) {
if (tree[p] != last)
ans[tree[p]] ++;
last = tree[p];
return;
}
if (l == r) {last = INF; return;}
int mid = (l + r) >> 1;
pushdown(p, l, r);
query(p << 1, l, mid, ql, qr);
query(p << 1 | 1, mid + 1, r, ql, qr);
}
int main () {
int n, T;
scanf("%d", &T);
while(T--) {
int tot = 0;
scanf("%d", &n);
if (n) {
for (int k = 0; k < n; k++) {
scanf("%d%d", L + k, R + k);
disc[tot++] = L[k];
disc[tot++] = R[k];
}
//离散化
sort(disc, disc + tot);
int size = (int)(unique(disc, disc + tot) - disc);
int temp = size;
for (int i = 1; i < temp; i++)
if (disc[i] - disc[i - 1] > 1)
disc[size++] = disc[i - 1] + 1;
sort(disc, disc + size);
for (int i = 0; i < n; i ++) {
L[i] = (int)(lower_bound(disc, disc + size, L[i]) - disc + 1);
R[i] = (int)(lower_bound(disc, disc + size, R[i]) - disc + 1);
}
memset(lazy, INF, sizeof(lazy));
memset(origin, INF, sizeof(origin));
memset(ans, 0, sizeof(ans));
last = INF;
build(1, 1, size);
for (int i = 0; i < n; i ++)
update(1, 1, size, L[i], R[i], i + 1);
query(1, 1, size, 1, size);
int cnt = 0;
for (int i = 1; i <= size; i ++)
if (ans[i]) cnt++;
printf("%dn", cnt);
}
else printf("0n");
}
}

Mayor's posters

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters and introduce the following rules:

Every candidate can place exactly one poster on the wall.
All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown).
The wall is divided into segments and the width of each segment is one byte.
Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections.
Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall.

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among the n lines contains two integer numbers l i and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= l i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l i, l i+1 ,... , ri.

Output

For each input data set print the number of visible posters after all the posters are placed.
The picture below illustrates the case of the sample input.