POJ3263 Tallest Cow差分模板练习

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本文解析了POJ3263题目，即牛视距问题，探讨了如何通过差分数组的方法确定每头牛的最大可能高度，以满足题目中给出的所有条件。介绍了使用差分技巧来解决区间更新问题的有效方法。

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POJ3263

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.

Sample Input
9 3 5 5
1 3
5 3
4 3
3 7
9 8
Sample Output
5
4
5
3
4
4
5
5
5

题意：有n头牛站成一行。如果两头牛之间的牛都比他们俩矮，那么这两头牛才可以互相看见。我们只知道第P头牛是最高的，高H。我们还知道M对关系，代表 $A_i$ $与B_i$ 这两头牛可以相互看见。求每头牛最大可能身高。

题解：
差分，建立一个数组 $D$ ， $D_i$ 代表从i开始有一个作用到最后，如果想要令这个作用只持续一段区间比如说[i, j]，那么从 $D_{j+1}$ 开始加上一个反作用即可。

复杂度 $O (n + m)$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;
typedef long long ll;

map<pair<int, int>, bool> existed;
int c[10010], d[10010];

int main(){
	int n, p, h, m;
	cin >> n >> p >> h >> m;
	for (int i = 1; i <= m; i++){
		int a, b;
		scanf("%d%d", &a, &b);
		if (a > b) swap(a, b);
		if(existed[make_pair(a, b)]) continue;
		d[a + 1]--, d[b]++, existed[make_pair(a, b)] = true;
	}
	for(int i = 1; i<= n; i++){
		c[i] = c[i - 1] + d[i];
		printf("%d\n", h + c[i]);
	}
	cout << endl;	
	return 0;
}