POJ2528 Mayor's posters 線段樹

POJ2528 Mayor's posters 線段樹

Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 49723   Accepted: 14419

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

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

首先想到離散化，正確的離散化的方法：
即每條線段用這條線段所覆蓋的單位區間的兩端的區間表示， 如圖中：線段①表示為：[1,4]；線段②表示為：[2,6]；線段③表示為：[3,4]。
離散後既對線段處理，我們可以把所以線段全部插入，然後再查詢一次。 即染色問題 維護這樣的信息 cover[k]:k對應的區間是否被同一種顏色完全染色。 顏色依次用1,2,3,...,n表示，初始是用0將區間全部 查詢時用一個bool型數組vis，查詢整個區間，遇見有什麼顏色，就把該顏色的vis設為true，最後 統計一下，vis裡true的個數即為答案。

#include 
#include 
#include 
using namespace std;

//其實離散後總共的單位區間最多可達到20000+19999個
//所以cover開到maxn*2*2*4
const int maxn=10000+10;
int cover[maxn*16],x[maxn],y[maxn],s[maxn*2];
unsigned short Hash[10000005];      //unsigned short才夠用，short不夠用，會re

void pushDown(int k){
    if(cover[k]>=0){
        cover[k*2]=cover[k*2+1]=cover[k];
        cover[k]=-1;
    }
}

void Insert(int a,int b,int c,int k,int l,int r){
    if(a<=l && r<=b){
        cover[k]=c;return ;
    }
    pushDown(k);
    int m=(l+r)/2;
    if(a<=m)
        Insert(a,b,c,k*2,l,m);
    if(b>m)
        Insert(a,b,c,k*2+1,m+1,r);
}

bool vis[maxn];

void ask(int k,int l,int r){
    if(cover[k]>=0){
        vis[cover[k]]=true;
        return ;
    }
    if(l==r) return ;
    pushDown(k);
    int m=(l+r)/2;
    ask(k*2,l,m);
    ask(k*2+1,m+1,r);
}

int main()
{
    int i,T,n;
    scanf(%d,&T);
    while(T--){
        scanf(%d,&n);
        memset(Hash,0,sizeof(Hash));
        int tot=0;
        for(i=0;i