题目描述

Farmer John's cows aregetting restless about their poor telephone service; they want FJ to replacethe old telephone wire with new, more efficient wire. The new wiring willutilize N (2 ≤ N ≤ 100,000) already-installed telephone poles, each with someheighti meters (1 ≤ heighti ≤ 100). The new wire will connect the tops of eachpair of adjacent poles and will incur a penalty cost C × the two poles' heightdifference 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 bemoved.

Farmer John figuresthat if he makes some poles taller he can reduce his penalties, though withsome other additional cost. He can add an integer X number of meters to a poleat a cost of X2.

Help Farmer Johndetermine the cheapest combination of growing pole heights and connecting wireso 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: Twospace-separated integers: N and C

* Lines 2..N+1: Linei+1 contains a single integer: heighti

输出格式：

* Line 1: The minimumtotal amount of money that it will cost Farmer John to attach the new telephonewire.

输入输出样例

输入样例#1： 复制

5 2
2
3
5
1
4

输出样例#1： 复制

15

算法分析：

设dp[i][j]为第i棵树的j高度的最小花费

 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));

但如下代码交上去超时：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
#define N 110005
int n,c,h[N],dp[N][105];
inline void read(int &n)
{
    char c='+';bool flag=0;n=0;
    while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
    while(c>='0'&&c<='9')n=n*10+c-48,c=getchar();
}
int main()  
{  
      
    while(scanf("%d%d",&n,&c)!=EOF)  
    { 
         int maxh=0;
         memset(dp,INT_INF,sizeof(dp));
         for(int i=1;i<=n;i++)
         {
            read(h[i]);
            maxh=max(h[i],maxh);
         }
         for(int i=h[1];i<=maxh;i++)
         {
            dp[1][i]=(i-h[1])*(i-h[1]);
            
         }
         for(int i=2;i<=n;i++)
           for(int j=h[i];j<=maxh;j++)
             for(int k=h[i-1];k<=maxh;k++)
         {
            dp[i][j]=min(dp[i][j],(j-h[i])*(j-h[i])+dp[i-1][k]+abs(j-k)*c);
            
         }
         
         int ans=INT_INF;
         for(int i=h[n];i<=maxh;i++)
            {ans=min(ans,dp[n][i]);}
         printf("%d",ans);
    }  
        return 0;  
}


所以需要优化，

dp[i][j]=min( (j-a[i])^2+f[i-1][k]+|j-k|*c )(j>=a[i])

当(j>=k) 时，dp[i][j]=(j-a[i])^2+j*c+min(f[i-1][k]-k*c) (j>=k)

  当(j<k)时， dp[i][j]= (j-a[i])^2-j*c+min(f[i-1][k]+k*c) (j<k)

设high[j]=min(dp[i-1][k]-k*c) (j>=k)

low[j]=min(dp[i-1][k]+k*c) (j<k)

方程就变为dp[i][j]=(j-a[i])^2+min(high[j]+j*c,low[j]-j*c);

AC代码

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
#define N 110005
int n,c,h[N],dp[N][105],low[105],high[105];
inline void read(int &n)
{
    char c='+';bool flag=0;n=0;
    while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
    while(c>='0'&&c<='9')n=n*10+c-48,c=getchar();
}
int main()  
{  
      
    while(scanf("%d%d",&n,&c)!=EOF)  
    { 
         int maxh=0;
         memset(dp,INT_INF,sizeof(dp));
          memset(low,INT_INF,sizeof(low));
           memset(high,INT_INF,sizeof(high));
         for(int i=1;i<=n;i++)
         {
            read(h[i]);
            maxh=max(h[i],maxh);
         }
          for(int i=1;i<=maxh;i++)
            dp[0][i]=0;
          
         for(int i=1;i<=n;i++)
         {
            
            for(int t=INT_INF, j=maxh;j>0;j--)
            {
                
                low[j]=t=min(t,dp[i-1][j]+c*j);
            }
            for(int t=INT_INF, j=1;j<=maxh;j++)
            {
                high[j]=t=min(t,dp[i-1][j]-c*j);
            }
            for(int j=h[i];j<=maxh;j++)
            dp[i][j]=(j-h[i])*(j-h[i])+min(high[j]+j*c,low[j]-j*c) ;
            
         }
         
         int ans=INT_INF;
         for(int i=h[n];i<=maxh;i++)
            {ans=min(ans,dp[n][i]);}
         printf("%d",ans);
    }  
        return 0;  
}