题目描述
FJ's N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.
FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form "cow 17 sees cow 34". This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.
For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.
有n头牛给定最高身高H和它的位置I，n头牛站成一列给定R个关系A，B 即A能看到B所以A------B之间所有牛小于AB且B的身高大于等于A
输入输出格式
输入格式：
Line 1: Four space-separated integers: N, I, H and R
Lines 2..R+1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.
输出格式：
Lines 1..N: Line i contains the maximum possible height of cow i.
输入输出样例
9 3 5 5 1 3 5 3 4 3 3 7 9 8
5 4 5 3 4 4 5 5 5
分析：
由题意可知在给定区间（A,B）中A,B除外所有元素均小于A,B所以可知各区间仅能包含或覆盖不能相交
——————————————————————
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反证法可证明：若区间(C,D)和区间（A,B)相交
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C D
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A B
由题意可得D高度小于A，B因而若C向右看一定先看到A>B与已知情况矛盾所以不成立
因此区间仅能相互覆盖所以问题转化为统计每个位置上区间覆盖次数
可用差分数组实现对于区间(A,B) ch[A+1]--;ch[B]++;
最后差分+最高高度H就是答案