「BZOJ1651」[Usaco2006 Feb] Stall Reservations 专用牛棚
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
有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求
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.
Sample Input
1 10
2 4
3 6
5 8
4 7
Sample Output
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.
HINT
不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3
题解
眼瞎了写了线段树，用差分序列就行了
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#include<iostream> #include<cstdio> using namespace std; 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; struct data{int l,r,mx,tag;}t[3000005]; void build(int k,int l,int r) { t[k].l=l;t[k].r=r; if(t[k].l==t[k].r)return; int mid=(l+r)>>1; build(k<<1,l,mid); build(k<<1|1,mid+1,r); } void pushdown(int k) { int l=t[k].l,r=t[k].r,tag=t[k].tag; if(l==r||!tag)return; t[k].tag=0; t[k<<1].mx+=tag;t[k<<1|1].mx+=tag; t[k<<1].tag+=tag;t[k<<1|1].tag+=tag; } void update(int k,int x,int y) { pushdown(k); int l=t[k].l,r=t[k].r; if(l==x&&r==y){t[k].mx++;t[k].tag++;return;} int mid=(l+r)>>1; if(mid>=y)update(k<<1,x,y); else if(mid<x)update(k<<1|1,x,y); else {update(k<<1,x,mid);update(k<<1|1,mid+1,y);} t[k].mx=max(t[k<<1].mx,t[k<<1|1].mx); } int main() { n=read(); build(1,1,1000000); for(int i=1;i<=n;i++) { int x=read(),y=read(); update(1,x,y); } pushdown(1); printf("%d",t[1].mx); return 0; } |
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