# bzoj 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚 -- 线段树

## 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB

## Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows. Help FJ by determining: * The minimum number of stalls required in the barn so that each cow can have her private milking period * An assignment of cows to these stalls over time

## Input

* Line 1: A single integer, N

* Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

## Output

* Line 1: The minimum number of stalls the barn must have.

* Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

5
1 10
2 4
3 6
5 8
4 7

## Sample Output

4

OUTPUT DETAILS:

Here's a graphical schedule for this output:

Time 1 2 3 4 5 6 7 8 9 10
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

## Source

#include<cstdio>
#define M 1000010
inline int read()
{
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x;
}
inline int max(int a,int b){return a>b?a:b;}
struct ljn{int l,r,maxx,lazy;}tr[4*M];
int n,x,y;
void make(int l,int r,int p)
{
tr[p].l=l,tr[p].r=r,tr[p].lazy=0;tr[p].maxx=0;
if(l==r) return;
int mid=(l+r)>>1;
make(l,mid,p<<1);
make(mid+1,r,p<<1|1);
}
void pd(int p)
{
int lz=tr[p].lazy;
tr[p<<1].lazy+=lz;
tr[p<<1|1].lazy+=lz;
tr[p<<1].maxx+=lz;
tr[p<<1|1].maxx+=lz;
tr[p].lazy=0;
}
void xg(int l,int r,int c,int p)
{
if(tr[p].l==l&&tr[p].r==r)
{
tr[p].lazy+=c;
tr[p].maxx+=c;
return;
}
pd(p);
int mid=(tr[p].l+tr[p].r)>>1;
if(mid>=r) xg(l,r,c,p<<1);
else if(mid<l) xg(l,r,c,p<<1|1);
else
{
xg(l,mid,c,p<<1);
xg(mid+1,r,c,p<<1|1);
}
tr[p].maxx=max(tr[p<<1].maxx,tr[p<<1|1].maxx);
}
int main()
{
n=read();
make(1,1000000,1);
for(int i=0;i<n;i++)
{
x=read();y=read();
xg(x,y,1,1);
}
printf("%d\n",tr[1].maxx);
}

#include<cstdio>
#define M 1000000
inline int max(int x,int y){return x>y?x:y;}
inline int read()
{
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x;
}
int n,a,b,t[M+10],ans,wei;
int main()
{
n=read();
while(n--)
{
a=read();b=read();
t[a]++;
t[b+1]--;
wei=max(b,wei);
}
for(int i=1;i<=wei;i++)
{
t[i]=t[i-1]+t[i];
ans=max(ans,t[i]);
}
printf("%d\n",ans);
}

posted @ 2017-01-17 14:24  lkhll  阅读(1990)  评论(0编辑  收藏  举报