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[Usaco2006 Feb]Stall Reservations 专用牛棚

【[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
5
1 10
2 4
3 6
5 8
4 7
Sample Output
4
//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
题解
差分思想
代码

#include using namespace std; int a[1000009],sum[1000009]; int i,j,n,b,c,max; int main() { scanf("%d",&n); for (i=1; i<=n; i++) { scanf("%d%d",&b,&c); a[b]+=1; a[c+1]-=1; } for (i=1; i<=1000000; i++) { sum[i]=sum[i-1]+a[i]; if (sum[i]>max) { max=sum[i]; } } printf("%d\n",max); return 0; }

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