http://poj.org/problem?id=2528
Mayor's posters
Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 29016Accepted: 8405Description
The citizens of Bytetown, AB,could not stand that the candidates in the mayoral electioncampaign have been placing their electoral posters at all places attheir whim. The city council has finally decided to build anelectoral wall for placing the posters and introduce the followingrules:
- Every candidate can place exactly one poster on thewall.
- All posters are of the same height equal to the height of thewall; 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 segmentis one byte.
- Each poster must completely cover a contiguous number of wallsegments.
They have built a wall 10000000 bytes long (such that there isenough place for all candidates). When the electoral campaign wasrestarted, the candidates were placing their posters on the walland their posters differed widely in width. Moreover, thecandidates started placing their posters on wall segments alreadyoccupied by other posters. Everyone in Bytetown was curious whoseposters will be visible (entirely or in part) on the last daybefore elections.
Your task is to find the number of visible posters when all theposters are placed given the information about posters' size, theirplace and order of placement on the electoral wall.
Input
The first line of input containsa number c giving the number of cases that follow. The first lineof data for a single case contains number 1 <= n<= 10000. The subsequent n lines describe theposters in the order in which they were placed. The i-th line amongthe n lines contains two integer numbers li and ri whichare the number of the wall segment occupied by the left end and theright 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 wallsegments numbered li, li+1 ,... , ri.
Output
For each input data set printthe number of visible posters after all the posters areplaced.
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
给你一个区间,然后用n条线段去覆盖,问最后能看到多少条线段.
思路:
由于线段的范围太大,而n小得多,我们可以考虑将区间按照位置离散化.也就是原来的1 20 30 40变为1 2 3 4
然后没条线段染上一种颜色,最后看有多少种颜色就可以了。。
#include
#include
#include
#include
using namespace std;
const int maxn=20010;
struct Tree
{
intld,rd;
intcol;
}T[maxn*4];
struct segment
{
intfrom;
intto;
}a[maxn*2];
int x[maxn*2],visCol;
bool hasCol[maxn*2];
void build(int ld,int rd,int i)
{
T[i].ld=ld;
T[i].rd=rd;
T[i].col=-1;//初始化为-1
if(ld+1==rd)
{
return;
}
intmid=(ld+rd)>>1;
build(ld,mid,i*2);
build(mid,rd,i*2+1);
}
void modify(int l,int r,int col,int i)
{
if(l<=T[i].ld&&r>=T[i].rd)
{
T[i].col=col;
return;
}
if(T[i].col!=-1)
{
T[i*2].col=T[i*2+1].col=T[i].col;
T[i].col=-1;
}
intmid=(T[i].ld+T[i].rd)>>1;
if(l>=mid)
{
modify(l,r,col,i*2+1);
}
elseif(r<=mid)
{
modify(l,r,col,i*2);
}
else
{
modify(l,mid,col,i*2);
modify(mid,r,col,i*2+1);
}
}
void query(int i)
{
if(T[i].col!=-1)
{
if(!hasCol[T[i].col])
{
hasCol[T[i].col]=true;
visCol++;
}
return;
}
if(T[i].ld+1==T[i].rd)
{
return;
}
query(i*2);
query(i*2+1);
}
int main()
{
intcases,n;
scanf("%d",&cases);
while(cases--)
{
scanf("%d",&n);
int cnt=0;
int i;
for(i=1;i<=n;i++)
{
scanf("%d%d",&a[i].from,&a[i].to);
++a[i].to;//不理解
x[cnt++]=a[i].from;//坐标保存在x【】数组里
x[cnt++]=a[i].to;
}
sort(x,x+cnt);//把x【】排序
cnt=unique(x,x+cnt)-x;//返回去重后x【】坐标的最后一个元素的下标
build(0,cnt-1,1);//建树
int leftPoint,rightPoint;//
for(i=1;i<=n;i++)
{
leftPoint=lower_bound(x,x+cnt,a[i].from)-x;
rightPoint=lower_bound(x,x+cnt,a[i].to)-x;