D - Mayor‘s posters
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 2528
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
//用C++提交WA，G++就能AC
//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=20011;
const int inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,n) for(int i=0;i<(n);i++)
template<class T>inline T read(T&x)
{
char c;
while((c=getchar())<=32);
bool ok=false;
if(c=='-')ok=true,c=getchar();
for(x=0; c>32; c=getchar())
x=x*10+c-'0';
if(ok)x=-x;
return x;
}
template<class T> inline void read_(T&x,T&y)
{
read(x);
read(y);
}
template<class T> inline void write(T x)
{
if(x<0)putchar('-'),x=-x;
if(x<10)putchar(x+'0');
else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
write(x);
putchar('\n');
}
//-------IO template------
typedef long long LL;
struct poster
{
int left,right;
}pos[maxn];
int x[maxn<<1];
int hash[10000011];
struct node
{
bool bcovered;
}p[maxn<<3];
///初始都可见，即false
void build(int rt,int L,int R)
{
p[rt].bcovered=false;
if(L==R)return ;
build(lson);
build(rson);
}
///插入一张海报，true是此张海报部分或者全部可见，false是不可见
bool update(int rt,int L,int R,int x,int y)
{
//printf("%d %d %d %d %d\n",rt,L,R,x,y);
if(p[rt].bcovered)return false;
if(L==x&&y==R)
{
return p[rt].bcovered=true;
}
bool ans;
if(y<=M)
ans=update(lson,x,y);
else if(x>M)
ans=update(rson,x,y);
else
{
ans=update(lson,x,M)|update(rson,M+1,y);//按位或，保证|的两边都会计算，不能用||，因为||是短路运算，如果左边是true，右边的就不会计算了
}
p[rt].bcovered=p[rt<<1].bcovered&&p[rt<<1|1].bcovered;//回溯回来更新父亲的状态
return ans;
}
int main()
{
//#ifndef ONLINE_JUDGE
// freopen("in.txt","r",stdin);
// #endif // ONLINE_JUDGE
int T;
int n,m,i,j,k,t;
read(T);
while(T--)
{
read(n);
int cnt=0;
For(i,n)
{
read_(pos[i].left,pos[i].right);
x[cnt++]=pos[i].left;
x[cnt++]=pos[i].right;
}
sort(x,x+cnt);
cnt=unique(x,x+cnt)-x;
int num=0;
For(i,cnt)///离散化
{
hash[x[i]]=num;
if(i<cnt-1)
{
if(x[i+1]-x[i]==1)//处理不相邻的离散化后依旧不相邻，相邻的依然相邻
num++;
else
num+=2;
}
}
build(1,0,num);
int ans=0;///从外往里贴海报（和真实的情况正好相反），保证后来的贴的对以前的没影响
for(i=n-1;i>=0;i--)if(update(1,0,num,hash[pos[i].left],hash[pos[i].right]))
ans++;
writeln(ans);
}
return 0;
}
/*
离散化要注意一点
1 10
1 3
6 10
应该变为
0 6
0 2
4 6
注意离散化前不相邻的，离散化后也是不相邻的
*/
D - Mayor's posters （线段树+离散化处理） POJ 2528