[BZOJ1651][Usaco2006 Feb]Stall Reservations 专用牛棚

1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB Submit: 990  Solved: 568 [Submit][Status][Discuss]

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


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.

 

问的就是被区间覆盖次数的最大值，用查分数组维护覆盖，然后扫一遍计算答案

#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;
}