NSOJ 4621 posters （离散化+线段树）

posters

时间限制: 1000ms

内存限制: 128000KB

64位整型: Java 类名:

题目描述

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.

输入

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 li 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 <= li <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered li, li+1 ,... , ri.

输出

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.
http://acm.pku.edu.cn/JudgeOnline/images/2528_1.jpg

样例输入

样例输出

来源

NYOJ

题意：

　　　在一面长墙上贴n张海报，第 i 张海报可以从 li 贴到 ri ,海报高低都相等（可以理解为一维模型），按照给出的方法依次粘贴，问最终还能看见海报有几张。

分析：

　　　起始点与终止点取值范围（1 <= i <= n, 1 <= li <= ri <= 10000000）然而我们用到的只有（1 <= n <= 10000）个，只用到了（1/1000）*2的点，要保存整个状态，开10000000的整形数组果断超内存，运用离散化思想用20000的整形数组便能保存整个状态。

　　　如果按照一般的方法，你在一张海报区域对应范围内逐个赋值保存状态，离散化后最坏的情况依然为2*c*n²果断超时，这时就要用到线段树来优化了，线段树用到二分思想故最坏情况为2*c*nlogn,(c的范围没有给出，索性考虑在内，但c应该也不会太大)。

 #include <iostream>

 #include <algorithm>

 #include <map>    //映射->离散化

 #include <set>  //自动去重

 using namespace std;

 #define MAXN 10005

 map <int,int> M;

 struct node{

     int left;

     int right;

     int index;

     node(): left(), right(), index() {}

 } Tree[MAXN * ], input[MAXN];

 int point[MAXN * ], sum;

 set <int> visible;

 void BuildTree(int root, int left, int right){ //建立线段树

     Tree[root].left = left;

     Tree[root].right = right;

     Tree[root].index = -;

     if (left +  < right){

         int mid = (left + right) / ;

         BuildTree(root * , left, mid);

         BuildTree(root *  + , mid, right);

     }

 }

 void UpdateTree(int k, int a, int b, int index){

     if (Tree[k].left == a && Tree[k].right == b){

         Tree[k].index = index;

         return;

     }

     if (Tree[k].index != -){

         Tree[ * k].index = Tree[ * k + ].index = Tree[k].index;

         Tree[k].index = -;

     }

     int mid = (Tree[k].left + Tree[k].right) / ;

     if (b <= mid)

         UpdateTree( * k, a, b, index);

     else if (a >= mid)

         UpdateTree( * k + , a, b, index);

     else {

         UpdateTree( * k, a, Tree[ * k].right, index);

         UpdateTree( * k + , Tree[ * k + ].left, b, index);

     }

 }

 void SearchTree(int k) { //便历

     if (Tree[k].index != -)

         visible.insert(Tree[k].index);

     else if (Tree[k].right > Tree[k].left + ) {

         SearchTree( * k);

         SearchTree( * k + );

     }

 }

 int main() {

     int c, n;

     cin >> c;

     while (c--) {

         cin >> n;

         sum = ;

         for (int i = ; i < n; i++) {

             cin >> input[i].left >> input[i].right;

             input[i].left--;

             point[sum++] = input[i].left;

             point[sum++] = input[i].right;

         }

         sort(point, point + sum);

         sum = unique(point, point + sum) - point;//去重后共sum个点

         M.clear();

         for(int i=;i<sum;i++){

             M[point[i]]=i; //映射

         }

         BuildTree(, , sum - );

         for (int i = ; i < n; i++) {

             UpdateTree(, M[input[i].left], M[input[i].right], i);

         }

         visible.clear();

         SearchTree();

         cout << visible.size() << endl;

     }

     return ;

 }