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
* Line 1: A single integer, N
* Lines 2..N+1: Line i+1 describes cow i‘s milking interval with two space-separated integers.
* 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.
#include <cstdio>
inline int readint(){
int n = 0;
char ch = getchar();
while(ch < ‘0‘ || ch > ‘9‘) ch = getchar();
while(ch <= ‘9‘ && ch >= ‘0‘){
n = (n << 1) + (n << 3) + ch - ‘0‘;
ch = getchar();
}
return n;
}
template <typename _Tp>
inline _Tp max_(const _Tp &a, const _Tp &b){
return a > b ? a : b;
}
int cnt[1000001] = {0}, n;
int main(){
n = readint();
int r = 0;
for(int s, t, i = 1; i <= n; i++){
s = readint();
t = readint();
cnt[s] ++;
cnt[t + 1] --;
r = max_(r, t);
}
int sum = 0, ans = 0;
for(int i = 1; i <= r; i++){
sum += cnt[i];
ans = max_(ans, sum);
}
printf("%d\n", ans);
return 0;
}
