P2885 [USACO07NOV]电话线Telephone Wire

题目描述

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.

给出若干棵树的高度，你可以进行一种操作：把某棵树增高h，花费为h*h。

操作完成后连线，两棵树间花费为高度差*定值c。

求两种花费加和最小值。

输入输出格式

输入格式：

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

输出格式：

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

输入输出样例

输入样例#1：

输出样例#1：15

一开始自己写了个DP，居然能过样例
特别兴奋，
但是交上去只有70分
发现时间复杂度有点高
思路比较简单：
我们可以很容易的看出这道题具有无后效性，
用dp[i][j]表示前i棵树，第i棵树高度为j的最小代价
先预处理一下dp[1][j]，然后对于每一棵树，我们枚举它的前一棵树的高度和这棵树的高度，
计算一下就好
时间复杂度n*h*h

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<cstdlib>
 9 using namespace std;
10 const int MAXN=100001;
11 const int INF =0x7f7f7f7f;
12 inline void read(int &n)
13 {
14     char c='+';bool flag=0;n=0;
15     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
16     while(c>='0'&&c<='9')n=n*10+c-48,c=getchar();
17 }
18 int dp[MAXN][201];// 第i棵树，高度为j的最小花费
19 int n,C; 
20 int a[MAXN];
21 int maxheight;
22 int main()
23 {
24     read(n);read(C);
25     memset(dp,INF,sizeof(dp));
26     for(int i=1;i<=n;i++)
27         read(a[i]),maxheight=max(maxheight,a[i]);
28     for(int i=a[1];i<=maxheight;i++)
29         dp[1][i]=(i-a[1])*(i-a[1]);
30         
31     for(int i=2;i<=n;i++)//枚举所有树 
32         for(int j=a[i];j<=maxheight;j++)//枚举这棵树的高度 
33             for(int k=a[i-1];k<=maxheight;k++)//枚举前一棵树的高度 
34                 dp[i][j]=min(dp[i][j],
35                             ((j-a[i])*(j-a[i])+(dp[i-1][k])+abs(j-k)*C));
36                             
37                             
38     int ans=0x7fffff;
39     for(int i=a[n];i<=maxheight;i++)
40         ans=min(ans,dp[n][i]);
41     printf("%d",ans);
42     return 0;
43 }

然后化简一下式子

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 using namespace std;
 6 const int MAXN=300005;
 7 const int INF =0x7fffff;
 8 const int maxheight=100;
 9 int dp[301];// 第i棵树，高度为j的最小花费
10 int f[301];
11 int n,C;
12 int a[MAXN];
13 int bgsum[MAXN];
14 int edsum[MAXN];
15 int main() {
16     scanf("%d%d",&n,&C);
17     for(int i=0; i<n; i++)
18         scanf("%d",&a[i]);
19     memset(dp, 0x3f, sizeof(dp));
20     for(int i=a[0]; i<=maxheight; i++)
21         dp[i]=(i-a[0])*(i-a[0]);
22 
23     for(int i=1; i<n; i++) { //枚举所有树
24         memcpy(f,dp,sizeof(dp));
25         for(int j=0; j<=maxheight; j++)    dp[j]=bgsum[j]=edsum[j]=INF;
26 
27         bgsum[0]=f[0];
28         for(int j=1; j<=maxheight; j++)
29             bgsum[j]=min(bgsum[j-1],f[j]-C*j);
30 
31         edsum[maxheight]=f[maxheight]+maxheight*C;
32         for(int j=maxheight-1; j>=0; j--)
33             edsum[j]=min(edsum[j+1],f[j]+C*j);
34 
35         for(int j=a[i]; j<=maxheight; j++) //枚举这棵树的高度
36             dp[j]=min(edsum[j]-C*j,bgsum[j]+C*j)+(j-a[i])*(j-a[i]);
37     }
38     int ans=0x7fffff;
39     for(int i=a[n-1]; i<=maxheight; i++)
40         ans=min(ans,dp[i]);
41     printf("%d",ans);
42     return 0;
43 }