【HDOJ】【2829】Lawrence

DP/四边形不等式

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　　做过POJ 1739 邮局那道题后就很容易写出动规方程：

　　　　dp[i][j]=min{dp[i-1][k]+w[k+1][j]}（表示前 j 个点分成 i 块的最小代价）

　　$w(l,r)=\sum_{i=l}^{r}\sum_{j=i+1}^{r}a[i]*a[j]$

　　那么就有 $w(l,r+1)=w(l,r)+a[j]*\sum\limits_{i=l}^{r}a[i]$

　　所以：w[i][j]明显满足 关于区间包含的单调性

　　然后我们大胆猜想，小(bu)心(yong)证明，w[i][j]满足四边形不等式，所以这题就跟邮局那题一样了……

　　咳咳好吧作为一个有节操的人，我还是尝试着证明了一下（结果发现用来证明的时间比我写代码的时间要长……）

　　先把w(i,j)的定义搬下来：\[ w(l,r)=\sum\limits_{i=l}^{r}\sum\limits_{j=i+1}^{r}a[i]*a[j] \]

　　形象一点来说就是：

　　　　对于$ i\leq i' < j \leq j' $

　　　　中间的都是要算两次，剩下的部分：

　　　　　　(左)表示w(i,i'-1)，[左]表示 $\sum_{k=i}^{i'-1}a[k] $

　　　　　　(中)表示w(i',j)，[中]表示 $\sum_{k=i'}^j a[k] $

　　　　　　(右)表示w(j+1,j')，[右]表示 $\sum_{k=j+1}^{j'} a[k] $

　　　　\[　w(i,j)+w(i',j')=(左)+[左]*[中]+(右)+[右]*[中]+(中) \\ w(i,j')+w(i',j)=(左+右)+[左+右]*[中]+(中)  \]

　　　　

　　　　其中\[ [左+右]*[中]=[左]*[中]+[右]*[中]  \]

　　　　但\[ (左+右)=(左)+(右)+[左]*[右] \]

　　　　所以\[ (左+右)>(左)+(右) \]

　　　　所以\[w(i,j)+w(i',j') \leq w(i,j')+w(i',j) \]　　

 //HDOJ 2829
 #include<cmath>
 #include<vector>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 #define pb push_back
 #define CC(a,b) memset(a,b,sizeof(a))
 using namespace std;
 int getint(){
     int v=,sign=; char ch=getchar();
     while(!isdigit(ch)) {if(ch=='-') sign=-; ch=getchar();}
     while(isdigit(ch))  {v=v*+ch-''; ch=getchar();}
     return v*sign;
 }
 const int N=,INF=~0u>>;
 const double eps=1e-;
 #define debug
 /*******************template********************/
 int dp[N][N],s[N][N],w[N][N],b[N],a[N],n,m;

 int main(){
     while(scanf("%d%d",&n,&m)!=EOF && n){
         m++;
         F(i,,n) a[i]=getint();
         F(i,,n){
             b[i]=a[i];
             w[i][i]=;
             F(j,i+,n){
                 w[i][j]=w[i][j-]+a[j]*b[i];
                 b[i]+=a[j];
             }
         }
         F(i,,n) F(j,,m) dp[j][i]=INF;
         F(i,,n){
             dp[][i]=w[][i];
             s[][i]=;
         }
         F(i,,m){
             s[i][n+]=n;
             D(j,n,i)
                 F(k,s[i-][j],s[i][j+])
                     if(dp[i-][k]+w[k+][j]<dp[i][j]){
                         s[i][j]=k;
                         dp[i][j]=dp[i-][k]+w[k+][j];
                     }
         }
         printf("%d\n",dp[m][n]);
     }
     return ;
 }

Lawrence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2448    Accepted Submission(s): 1093

Problem Description

T.
E. Lawrence was a controversial figure during World War I. He was a
British officer who served in the Arabian theater and led a group of
Arab nationals in guerilla strikes against the Ottoman Empire. His
primary targets were the railroads. A highly fictionalized version of
his exploits was presented in the blockbuster movie, "Lawrence of
Arabia".

You are to write a program to help Lawrence figure out
how to best use his limited resources. You have some information from
British Intelligence. First, the rail line is completely linear---there
are no branches, no spurs. Next, British Intelligence has assigned a
Strategic Importance to each depot---an integer from 1 to 100. A depot
is of no use on its own, it only has value if it is connected to other
depots. The Strategic Value of the entire railroad is calculated by
adding up the products of the Strategic Values for every pair of depots
that are connected, directly or indirectly, by the rail line. Consider
this railroad:

Its Strategic Value is 4*5 + 4*1 + 4*2 + 5*1 + 5*2 + 1*2 = 49.

Now,
suppose that Lawrence only has enough resources for one attack. He
cannot attack the depots themselves---they are too well defended. He
must attack the rail line between depots, in the middle of the desert.
Consider what would happen if Lawrence attacked this rail line right in
the middle:

The Strategic Value of the remaining railroad is 4*5 + 1*2 = 22. But, suppose Lawrence attacks between the 4 and 5 depots:

The Strategic Value of the remaining railroad is 5*1 + 5*2 + 1*2 = 17. This is Lawrence's best option.

Given
a description of a railroad and the number of attacks that Lawrence can
perform, figure out the smallest Strategic Value that he can achieve
for that railroad.

 

Input

There
will be several data sets. Each data set will begin with a line with
two integers, n and m. n is the number of depots on the railroad
(1≤n≤1000), and m is the number of attacks Lawrence has resources for
(0≤m<n). On the next line will be n integers, each from 1 to 100,
indicating the Strategic Value of each depot in order. End of input will
be marked by a line with n=0 and m=0, which should not be processed.

 

Output

For
each data set, output a single integer, indicating the smallest
Strategic Value for the railroad that Lawrence can achieve with his
attacks. Output each integer in its own line.

 

Sample Input

4 1
4 5 1 2
4 2
4 5 1 2
0 0

 

Sample Output

17
2

 

Source

2009 Multi-University Training Contest 2 - Host by TJU