poj 2528 Mayor's posters（扫描线+堆维护||离散化+线段树）
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Mayor's posters
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. 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 Source
Alberta Collegiate Programming Contest 2003.10.18
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题目：http://poj.org/problem?id=2528
题意：给你一块很长的墙，不断的在墙上的一些区间贴广告，可以覆盖，问，最后可以看到的广告有几条。。。
分析：这题可以用扫描线+堆维护，也就是从左到右扫描，遇到广告的左端就把广告加到堆里面，保存的信息是第几次贴的广告，遇到右端就把广告从堆里面删掉就行，在删掉前，堆顶元素就是可见的，标记下。。。最后枚举求和
另一种做法就是线段树，因为坐标范围太大，需要离散化，也就是把普通线段树的坐标，换成数组里的值，也就是不连续的坐标，然后就是普通的线段树成段更新了，也是挺好写的，具体看代码吧
扫描线+堆维护 代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
const int mm=11111;
struct node
{
int h,id;
}g[mm<<1];
int p[mm],flag[mm],vis[mm],heap[mm];
int sz;
void upsift(int k)
{
int i=k,j;
while(i>1)
{
j=i>>1;
if(heap[i]>heap[j])
{
heap[0]=heap[i],heap[i]=heap[j],heap[j]=heap[0];
p[heap[i]]=i,p[heap[j]]=j;
i=j;
}
else break;
}
}
void downsift(int k)
{
int i=k,j;
while(i*2<=sz)
{
j=i<<1;
if(j<sz&&heap[j+1]>heap[j])++j;
if(heap[i]<heap[j])
{
heap[0]=heap[i],heap[i]=heap[j],heap[j]=heap[0];
p[heap[i]]=i,p[heap[j]]=j;
i=j;
}
else break;
}
}
bool cmp(node a,node b)
{
return a.h<b.h;
}
int main()
{
int i,j,l,r,n,m,t,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(m=0,i=1;i<=n;++i)
{
scanf("%d%d",&l,&r);
g[m].h=l,g[m].id=i,++m;
g[m].h=r,g[m].id=i,++m;
}
sort(g,g+m,cmp);
sz=i=0;
memset(flag,0,sizeof(flag));
memset(vis,0,sizeof(vis));
while(i<m)
{
r=i+1;
while(r<m&&g[r].h==g[i].h)++r;
for(j=i;j<r;++j)
{
if(!flag[g[j].id])
{
++sz;
heap[sz]=g[j].id;
p[g[j].id]=sz;
upsift(p[g[j].id]);
downsift(p[g[j].id]);
}
++flag[g[j].id];
}
vis[heap[1]]=1;
for(j=i;j<r;++j)
if(flag[g[j].id]>1)
{
heap[p[g[j].id]]=l=heap[sz--];
p[l]=p[g[j].id];
upsift(p[l]);
downsift(p[l]);
flag[g[j].id]=0;
}
i=r;
}
for(ans=i=0;i<=n;++i)ans+=vis[i];
printf("%d\n",ans);
}
return 0;
}
离散化+线段树 代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int mm=22222;
int dly[mm<<2];
int l[mm],r[mm],a[mm],vis[mm];
void pushdown(int rt)
{
dly[rt<<1]=dly[rt<<1|1]=dly[rt];
dly[rt]=0;
}
void build()
{
memset(dly,0,sizeof(dly));
}
void updata(int L,int R,int id,int l,int r,int rt)
{
if(L<=a[l]&&R>=a[r])
{
dly[rt]=id;
return;
}
if(dly[rt])pushdown(rt);
int m=(l+r)>>1;
if(a[m]>=L)updata(L,R,id,lson);
if(m<r&&a[m+1]<=R)updata(L,R,id,rson);
}
void query(int rt)
{
if(dly[rt])
{
vis[dly[rt]]=1;
return;
}
query(rt<<1);
query(rt<<1|1);
}
int main()
{
int i,j,t,n,m,ans;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(m=0,i=1;i<=n;++i)
{
scanf("%d%d",&l[i],&r[i]);
a[m++]=l[i];
a[m++]=r[i];
}
sort(a,a+m);
for(j=i=0;i<m;++i)
if(a[i]!=a[j])a[++j]=a[i];
m=j;
build();
memset(vis,0,sizeof(vis));
for(i=1;i<=n;++i)
updata(l[i],r[i],i,0,m,1);
query(1);
for(ans=i=0;i<=n;++i)ans+=vis[i];
printf("%d\n",ans);
}
return 0;
}