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

>原题链接<

## 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

4

## 思路：

对每次修改差分，对最后差分求前缀和既是原数组，求一个max即可

#include <cstdio>
#define max(a,b) ((a)>(b)?(a):(b))
int c;
int main() {
int n,i,x,y,m=0,ans=0;
scanf("%d",&n);
for(i=1;i<=n;i++) {
scanf("%d%d",&x,&y);
m=max(m,y);
c[x]++;
c[y+1]--;
}
x=0;
for(i=1;i<=m;i++) {
x+=c[i];
ans=max(x,ans);
}
printf("%d\n",ans);
}


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posted @ 2018-05-28 20:53  TOBICHI_ORIGAMI  阅读(23)  评论(0编辑  收藏  举报