poj2528

Mayor's posters

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 26755 Accepted: 7725

Description

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. 

Sample Input

151 42 68 103 47 10

Sample Output

4

Source

Alberta Collegiate Programming Contest 2003.10.18

如果每个叶子节点都代表一块瓷砖,那么线段树会导致MLE,即单位区间的数目太多。需要用离散化方法,将所有海报的端点瓷砖排序,把每个海报的端点瓷砖都看做一个单位区间,两个相邻的端点瓷砖之间的部分是一个单位区间这样最多会有20000 + 19999个单位区间。关键: 插入数据的顺序 ------ 从上往下依次插入每张海报,这样后插入的海报不可能覆盖先插入的海报,因此插入一张海报时,如果发现海报对应区间有一部分露出来,就说明该海报部分可见。 

#include <iostream> #include <algorithm> #include <math.h> using namespace std;int n;struct CPost {    int L, R;};CPost posters[10100];int x[20200]; //海报的端点瓷砖编号int hash[10000010]; //hash[i]表示瓷砖i所处的离散化后的区间编号 struct CNode {    int L, R;    bool bCovered; //区间[L,R]是否已经被完全覆盖     CNode * pLeft, * pRight;};CNode Tree[1000000];int nNodeCount = 0;int Mid(CNode * pRoot) {    return (pRoot->L + pRoot->R) / 2;}void BuildTree(CNode * pRoot, int L, int R) {    pRoot->L = L;    pRoot->R = R;    pRoot->bCovered = false;    if (L == R)        return;    nNodeCount++;    pRoot->pLeft = Tree + nNodeCount;    nNodeCount++;    pRoot->pRight = Tree + nNodeCount;    BuildTree(pRoot->pLeft, L, (L + R) / 2);    BuildTree(pRoot->pRight, (L + R) / 2 + 1, R);}bool Post(CNode *pRoot, int L, int R) { //插入一张正好覆盖区间[L,R]的海报,返回true则说明区间[L,R]是部分或全部可见的    if (pRoot->bCovered) return false;    if (pRoot->L == L && pRoot->R == R) {        pRoot->bCovered = true;        return true;    }    bool bResult;    if (R <= Mid(pRoot))        bResult = Post(pRoot->pLeft, L, R);    else if (L >= Mid(pRoot) + 1)        bResult = Post(pRoot->pRight, L, R);    else {        bool b1 = Post(pRoot->pLeft, L, Mid(pRoot));        bool b2 = Post(pRoot->pRight, Mid(pRoot) + 1, R);        bResult = b1 || b2;    }    //要更新根节点的覆盖情况    if (pRoot->pLeft->bCovered && pRoot->pRight->bCovered)        pRoot->bCovered = true;    return bResult;}int main() {    int t;    int i, j, k;    scanf("%d", &t);    int nCaseNo = 0;    while (t--) {        nCaseNo++;        scanf("%d", &n);        int nCount = 0;        for (i = 0; i < n; i++) {            scanf("%d%d", & posters[i].L, & posters[i].R);            x[nCount++] = posters[i].L;            x[nCount++] = posters[i].R;        }        sort(x, x + nCount);        nCount = unique(x, x + nCount) - x; //去掉重复元素        //下面离散化        int nIntervalNo = 0;        for (i = 0; i < nCount; i++)    {            hash[x[i]] = nIntervalNo;            if (i < nCount - 1) {                if (x[i + 1] - x[i] == 1) nIntervalNo++;                else                    nIntervalNo += 2;            }        }        BuildTree(Tree, 0, nIntervalNo);        int nSum = 0;        for (i = n - 1; i >= 0; i--) // 从后往前看每个海报是否可见            if (Post(Tree, hash[posters[i].L], hash[posters[i].R]))                nSum++;        printf("%d\n", nSum);    }    return 0;}