POJ3612Telephone Wire题解动态规划DP

最新推荐文章于 2018-11-30 20:04:25 发布

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本文介绍了一种通过调整电线杆高度来最小化电话线安装总成本的方法。利用动态规划技术，详细展示了如何优化成本，避免因电线杆高度不同而产生的额外费用，并提供了具体的实现代码。

Telephone Wire

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1735		Accepted: 596

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 height_i meters (1 ≤ height_i ≤ 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: height_i

Output

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

Sample Input

Sample Output

Source

USACO 2007 November Gold

状态：

d[i][j]表示第i个高度为j时最小花费

状态转移方程：

d[i][j]=min(d[i][j],d[i-1][k]+abs(j-k)*m+(a[i]-j)*(a[i]-j));

这样肯定超时，不知道怎么优化

看了别人的解题报告:

/*
DP
设f[i][j]表示第i个电线杆高度为j时的最小花费,
DP方程为f[i][j]=min( (j-a[i])^2+f[i-1][k]+|j-k|*c )
直接根据这条方程做的话复杂度是100000*100*100,明显会超时.
可以将方程进行化简：
f[i][j]=min( (j-a[i])^2+f[i-1][k]+|j-k|*c ) (j>=a[i])
=(j-a[i])^2+j*c+min(f[i-1][k]-k*c) (j>=k)
(j-a[i])^2-j*c+min(f[i-1][k]+k*c) (j<k)
设high[j]=min(f[i-1][k]-k*c) (j>=k)
low[j]=min(f[i-1][k]+k*c) (j<k)
方程就变为f[i][j]=(j-a[i])^2+min(high[j]+j*c,low[j]-j*c);
可以通过预处理计算先出high[j],low[j].
*/

代码：

#include<cstdio> #define min(a,b) ((a)<(b)?(a):(b)) #define inf 1<<30 int main() { int n,m,i,j,t,ans,a,low[105],high[105],d[105]; while(scanf("%d%d",&n,&m)==2) { scanf("%d",&a); for(i=1;i<=100;i++) d[i]=i<a?inf:(a-i)*(a-i); for(i=1;i<n;i++) { scanf("%d",&a); for(t=inf,j=100;j>0;j--) low[j]=t=min(t,d[j]+j*m); for(t=inf,j=1;j<=100;j++) { high[j]=t=min(t,d[j]-j*m); d[j]=inf; } for(j=a;j<=100;j++) d[j]=(j-a)*(j-a)+min(low[j]-j*m,high[j]+j*m); } for(ans=inf,i=1;i<=100;i++) ans=min(ans,d[i]); printf("%d/n",ans); } }