关于DP的优化方法有很多种，低级的有矩阵快速幂，高级一点的比如四边形不等式优化、斜率优化等等。

因为在动态规划中，有这样的一类问题

状态转移方程 dp[i][j]=min{dp[i][k-1]+dp[k][j]}+w[i][j]  （k>i&&k<=j）  时间复杂度为 O(n^3)

且有如下一些定义和定理：

如果一个函数w[i][j]，满足 w[i][j]+w[i'][j']<=w[i][j']+w[i'][j]（ i<=i'<=j<=j'） 则称w满足凸四边形不等式

如果一个函数w[i][j]，满足 w[i'][j]<=w[i][j'] （ i<=i'<=j<=j' ）则称w关于区间包含关系单调

 

定理1：如果w同时满足四边形不等式和区间单调关系，则dp也满足四边形不等式

 

定理2：如果定理1条件满足时让dp[i][j]取最小值的k为K[i][j]，则K[i][j-1]<=K[i][j]<=K[i+1][j]

注：定理2是四边形不等式优化的关键所在，它说明了决策量具有单调性，然后我们可以据此来缩小决策枚举的区间，进行优化

 

定理3：w为凸当且仅当 w[i][j]+w[i+1][j+1]<=w[i+1][j]+w[i][j+1]

         注：定理3其实就是验证w是否为凸的方法，就是固定一个变量，然后看成是一个一元函数，进而判断单调性。

如，我们可以固定j算出w[i][j+1]-w[i][j]关于i的表达式，看它是关于i递增还是递减，如果是递减，则w为凸

 

 

 

   以上三个定理来自于黑书，具体证明过程太啰嗦，各种乱走符号，所以懒得贴上来，记住结论会用就差不多了。

   这种优化方法的一般步骤是：

先证明w[i][j+1]-w[i][j]关于i的表达式的单调性，如果递减，则w满足凸四边形不等式，再证明w是否同时满足区间关系单调性，

如果两条都满足，则推出dp也满足凸四边形不等式，所以状态转移方程dp[i][j]=min{dp[i][k-1]+dp[k][j]}+w[i][j]  （k>i&&k<=j）中的决策量s[i][j]（也就是k）满足s[i][j-1]<=s[i][j]<=s[i+1][j]，因此s[i][j]的枚举区间由(i+1,j)缩小为(s[i][j-1],s[i+1,j])，使复杂度从O(n^3)下降到O(n^2)。

 

PS：实际操作中，我们往往并不需要进行烦躁的证明，而只需要打表，然后观察就行了

如w[i][j],dp[i][j]是否满足四边形不等式啊，w[i][j]是否单调啊，决策函数K[i][j]是否满足定理2的不等式关系啊，都可以通过打表来搞

 

 

 

抽象的理论太烦躁，看一下具体的例子。。。。

                                                          

                                                                                  HDU2829

                                    Lawrence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2202    Accepted Submission(s): 965

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

 

 

 

 

容易求出dp方程：f[m][n] = Min{f[m-1][k]+w[k+1][n]}

令w[i][j](i<=j)表示(a[i]*a[i+1]+a[i]*a[i+2]+....+a[i]*a[j])+(a[i+1]*a[i+2]+a[i+1]=a[i+3]+....)+....+a[j-1]*a[j]——(1)

则(1)式可以表示为[(a[i]+a[i+1]+...+a[j])*(a[i]+a[i+1]+...+a[j])-a[i]^2-a[i+1]^2-....-a[j]^2] / 2

即(1) = [(sum1[j]-sum1[i-1])^2-(sum2[j]-sum2[i-1])]/2。

其中sum1[i]表示从0到i元素的和，sum2[i]表示从0到i元素的平方的和，可以用O(n)的时间预处理。

现在我们固定j，看一看函数 w[i][j+1]-w[i][j] 随i增加的增减性，很明显如果i加1的话，w[i][j+1]比w[i][j]减少的要多，所以该函数递减，所以w为凸；

而区间包含关系是很显然的，所以w函数满足四边形不等式，所以可以推得f[m][n]也满足四边形不等式。

决策变量范围：s[m-1][n]<=s[m][n]<=s[m][n+1]。

 

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 usingnamespace std; 5  6 constint N =1010; 7  8 typedef longlong llg; 9 10 int n, m, a[N], sum1[N], sum2[N], s[N][N];11 llg f[N][N];12 13 void dp()14 {15     int i, j, k, a, b;16     llg tmp;17     memset(f, -1, sizeof(f));18     for(i =0; i <= n; i++)19     {20         tmp = sum1[i];21         f[0][i] = (tmp*tmp - sum2[i]) /2;22         s[0][i] =0;23     }24     for(i =1; i <= m; i++)25         for(j = n; j >= i; j--)26         {27             s[i][n+1] = n;28             a = s[i-1][j];29             b = s[i][j+1];30             for(k = a; k <= b; k++)31             {32                 tmp = sum1[j] - sum1[k];33                 tmp = f[i-1][k]+(tmp*tmp - (sum2[j]-sum2[k])) /2;34                 if(f[i][j]==-1|| f[i][j]>tmp)35                 {36                     f[i][j] = tmp;37                     s[i][j] = k;38                 }39             }40         }41 }42 43 int main()44 {45     int i;46     while(scanf("%d%d", &n, &m) != EOF)47     {48         if(n==0&& m==0)  break;49         sum1[0] = sum2[0] =0;50         for(i =1; i <= n; i++)51         {52             scanf("%d", a+i);53             sum1[i] = sum1[i-1] + a[i];54             sum2[i] = sum2[i-1] + a[i]*a[i];55         }56         dp();57         printf("%lld\n", f[m][n]);58     }59     return0;60 }