Problem Description

Farmer John's cows are getting restless about their poor telephone service; they want FJ to replace the old telephone wire with new, more efficient wire. The new wiring will utilize N (2 ≤ N ≤ 100,000) already-installed telephone poles, each with some heighti meters (1 ≤ heighti ≤ 100). The new wire will connect the tops of each pair of adjacent poles and will incur a penalty cost C × the two poles' height difference for each section of wire where the poles are of different heights (1 ≤ C ≤ 100). The poles, of course, are in a certain sequence and can not be moved.
Farmer John figures that if he makes some poles taller he can reduce his penalties, though with some other additional cost. He can add an integer X number of meters to a pole at a cost of X2.
Help Farmer John determine the cheapest combination of growing pole heights and connecting wire so that the cows can get their new and improved service.

Input

Line 1: Two space-separated integers: N and C
Lines 2..N+1: Line i+1 contains a single integer: heighti

Output

Line 1: The minimum total amount of money that it will cost Farmer John to attach the new telephone wire.

Sample Input

5 2

2
3
5
1
4

Sample Output

15

題意：要改造n個電線桿，相鄰兩電線桿改造費用=C*兩電線桿的高度差，也可對任意電線桿進行加長x，但要花費加長費用x^2，求改造電線桿的最少的費用。

思路

用f[i][j]表示第i棵樹高度為j的時候的最小代價，枚舉相鄰兩棵樹高度即可，狀態方程：f[i][j]=min(f[i][k]+abs(j-k)+(j-a[i])^2)，測試后發現超時，不知道怎么優化，看了他人題解。

對狀態方程進行優化，利用分類討論j<k和j>=k，將k維度用兩個數組預處理，即：high[j]=min(f[i-1][k]-k*c) (j>=k)，low[j]=min(f[i-1][k]+k*c) (j<k)，原狀態方程就變為：f[i][j]=(j-a[i])^2+min(high[j]+j*c,low[j]-j*c);

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 105 #define MOD 100001 #define E 1e-12 using namespace std; int low[N],high[N],f[N]; int main() {int n,m;while(scanf("%d%d",&n,&m)!=EOF){int a;scanf("%d",&a);for(int i=1;i<=100;i++){if(i<a)f[i]=INF;elsef[i]=(a-i)*(a-i);}for(int i=1;i<n;i++){int temp=INF;scanf("%d",&a);for(int j=100;j>0;j--)//對low預處理{temp=min(temp,f[j]+j*m);low[j]=temp;}for(int j=1;j<=100;j++)//對high預處理{temp=min(temp,f[j]-j*m);high[j]=temp;f[j]=INF;}for(int j=a;j<=100;j++)f[j]=(j-a)*(j-a)+min(low[j]-j*m,high[j]+j*m);}int ans=INF;for(int i=1;i<=100;i++)ans=min(ans,f[i]);printf("%d\n",ans);} }

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總結

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