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:
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.
- 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.
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
题目大意：在一块儿宣传栏中贴宣传单，规定宽度相同，长度不同。按一定顺序贴，给出宣传单的起始和结束位置，这里位置不能忽略成点（而是一段长度），问最后会看到几个宣传单。
解题思路：由于给出的数据范围多大，所以要先进行离散化减少复杂度，因为这里给出的不是”点“是带长度的，所以一般的离散化会出现离散失真，这里可以在离散的时候增加技巧，即在不相邻的数据里增加分隔点，凸显不连续。然后进行线段树成段替换，最后求出整个区间的不同宣传单个数。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int maxn=50000;
int lm[maxn/2],rm[maxn/2];
int col[maxn*4];
int Hash[maxn];
int A[maxn];
int num=0;
int discretization(int l,int r,int key){ //离散化
while(l<=r){
int m=(l+r)/2;
if(key==A[m]){
return m;
}else if(key<A[m]){
r=m-1;
}else{
l=m+1;
}
}
}
void PushDown(int rt){
if(col[rt]!=-1){
col[rt*2]=col[rt];
col[rt*2+1]=col[rt];
col[rt]=-1;
}
}
void update(int rt,int L,int R,int l_ran,int r_ran,int _col){
if(l_ran<=L&&R<=r_ran){
col[rt]=_col;
return ;
}
PushDown(rt);
if(l_ran<=mid)
update(lson,l_ran,r_ran,_col);
if(r_ran>mid)
update(rson,l_ran,r_ran,_col);
}
void query(int rt,int L,int R){
if(col[rt]!=-1){
if(!Hash[col[rt]]){
num++;
Hash[col[rt]]=1;
}
return ;
}
if(L==R)
return ;
query(lson);
query(rson);
}
void debug(){
for(int i=1;i<32;i++){
printf("%d %d
",i,col[i]);
}
}
int main(){
int t;
scanf("%d",&t);
while(t--){
int n,nn=0,m;
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d%d",&lm[i],&rm[i]);
A[nn++]=lm[i];
A[nn++]=rm[i];
}
sort(A,A+nn);
m=1;
for(int i=0;i<nn-1;i++){ //去重
if(A[i]!=A[i+1]){
A[m++]=A[i+1];
}
}
for(int i=m-1;i>0;i--){ //添加分隔点
if(A[i]!=A[i-1]+1){
A[m++]=A[i-1]+1;
}
}
sort(A,A+m);
int tml,tmr;
memset(col,-1,sizeof(col));
for(int i=0;i<n;i++){
tml= discretization(0,m-1,lm[i]); //离散化
tmr= discretization(0,m-1,rm[i]); //离散化
update(1,0,m-1,tml,tmr,i);
}
// debug();
memset(Hash,0,sizeof(Hash));
num=0;
query(1,0,m-1);
printf("%d
",num);
}
return 0;
}