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Description
The Queen of England has n trees growing in a row in her garden. At that, the i-th (1 ≤ i ≤ n) tree from the left has height ai meters. Today the Queen decided to update the scenery of her garden. She wants the trees‘ heights to meet the condition: for all i(1 ≤ i < n), ai + 1 - ai = k, where k is the number the Queen chose.
Unfortunately, the royal gardener is not a machine and he cannot fulfill the desire of the Queen instantly! In one minute, the gardener can either decrease the height of a tree to any positive integer height or increase the height of a tree to any positive integer height. How should the royal gardener act to fulfill a whim of Her Majesty in the minimum number of minutes?
Input
The first line contains two space-separated integers: n, k (1 ≤ n, k ≤ 1000). The second line contains n space-separated integersa1, a2, ..., an (1 ≤ ai ≤ 1000) — the heights of the trees in the row.
Output
In the first line print a single integer p — the minimum number of minutes the gardener needs. In the next p lines print the description of his actions.
If the gardener needs to increase the height of the j-th (1 ≤ j ≤ n) tree from the left by x(x ≥ 1) meters, then print in the corresponding line "+ j x". If the gardener needs to decrease the height of the j-th (1 ≤ j ≤ n) tree from the left by x(x ≥ 1) meters, print on the corresponding line "- j x".
If there are multiple ways to make a row of trees beautiful in the minimum number of actions, you are allowed to print any of them.
Sample Input
4 1
1 2 1 5
2
+ 3 2
- 4 1
4 1
1 2 3 4
0
题意：给定n棵树，每棵树的高度为a[i]，每次可将一棵树砍掉或增加任意高度，给定一个整数k，求砍或增的最少次数使得树的高度满足a[i+1]-a[i]=k（即斜率为k）
思路：由于数据是1000，所以直接暴力，枚举任意序列中的数对a[i]和a[j]组成的直线，选择a[k]中在该直线上最多的直线，然后修改树的高度。复杂度o(n^3).
然而如果数据改为10^6呢，当然有更神奇的o(n)解法.
将每个数减去k*(j-1)，即将所求的斜直线变为横直线，这样求经过点最多的横直线只需扫一遍数组，用map记录下每个横直线经过的点数，再扫一遍找出最优的横直线，再扫一遍即可算出答案了。复杂度o(n)！！！
然后这里有一点比较坑的是树的高度不能出现负数.
下面是o(n)的代码：
#include<bits/stdc++.h> using namespace std; const int maxn=1000100; const int INF=(1<<29); typedef long long ll; int n,k; int a[maxn]; map<int,int> mp; int main() { while(cin>>n>>k){ for(int i=1;i<=n;i++) scanf("%d",&a[i]); mp.clear(); for(int i=1;i<=n;i++) a[i]-=(i-1)*k; for(int i=1;i<=n;i++) mp[a[i]]++; int Max=-INF,x=1; for(int i=1;i<=n;i++) if(mp[a[i]]>Max&&a[i]>0) Max=mp[x=a[i]]; int cnt=0; for(int i=1;i<=n;i++) if(a[i]!=x) cnt++; cout<<cnt<<endl; for(int i=1;i<=n;i++){ if(a[i]!=x){ if(x-a[i]>0) printf("+ %d %d\n",i,x-a[i]); else printf("- %d %d\n",i,a[i]-x); } } } return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4528773.html
