Tallest Cow（差分，前缀和）

最新推荐文章于 2022-10-05 21:55:40 发布

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最新推荐文章于 2022-10-05 21:55:40 发布

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分类专栏：差分前缀和文章标签：算法

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解决一个涉及牛群高度的秘密问题，通过差分算法确定每头牛的最大可能高度，确保所有给定条件正确无误。

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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.

输入描述：

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.

输出描述：

Lines 1…N: Line i contains the maximum possible height of cow i.

样例：

输入：

9 3 5 5
1 3
5 3
4 3
3 7
9 8

输出：

5
4
5
3
4
4
5
5
5

思路：

对区间进行操作，很明显可以使用差分做。
a看到b，相当于对a-b中间的高度减一，差分数组表示为cf[a+1]- -, cf[b]++;
第一个的高度为H（因为求最高的时候全部的高度初始化为H），接下来进行前缀和求每个牛对应的高度并且输出即可。

代码：

#include<stdio.h>
#include<map>
#include<iostream>
using namespace std;
int cf[1000009], res, x, y;
int N, I, H, R;
map<pair<int,int>, bool>mp;
int main(){
	cin>>N>>I>>H>>R;
	for(int i=0; i<R; i++){
		cin>>x>>y;
		if(x>y) swap(x,y);
		if(mp[make_pair(x,y)]) continue;
		cf[x+1]--;
		cf[y]++;
		mp[make_pair(x,y)]=true;
	}
	for(int i=1; i<=N; i++){
		res+=cf[i];
		cout<<H+res<<endl;
	}
	return 0;
}