HDU 3516

Tree Construction

Problem Description

Consider a two-dimensional space with a set of points (xi, yi) that satisfy xi < xj and yi > yj for all i < j. We want to have them all connected by a directed tree whose edges go toward either right (x positive) or upward (y positive). The figure below shows an example tree.


Write a program that finds a tree connecting all given points with the shortest total length of edges.

 

Input

The input begins with a line that contains an integer n (1 <= n <= 1000), the number of points. Then n lines follow. The i-th line contains two integers xi and yi (0 <= xi, yi <= 10000), which give the coordinates of the i-th point.

 

Output

Print the total length of edges in a line.

 

Sample Input

51 52 43 34 25 1110000 0

 

Sample Output

120

 

一道典型的四边形不等式优化。

当f[i,j]满足f[i,j]+f[i',j']<=f[i',j]+f[i,j'],i<=i'<=j<=j'时，称f满足四边形不等式；

当f[i,j]满足f[i',j]<=f[i,j']时，称f关于区间包含关系单调。

如果状态转移方程为m[i,j]=min{m[i,k-1]+m[k,j]|i<k<=j}+w[i,j],

且w[i,j]为区间包含关系的单调函数，且满足四边形不等式，则m[i,j]也满足四边形不等式，

进一部推出s[i,j-1]<=s[i,j]<=s[i+1,j]。

于是m[i,j]=min{m[i,k-1],m[k,j]|s[i,j-1]<=k<=s[i+1,j]}+w[i,j]。

证明附上国家队毛子青的论文:ACM国家集训队论文集(依据内容)\5、动态规划\《动态规划算法的优化技巧》(由于不能上传文件，将论文放在了百度云上)。

http://pan.baidu.com/s/1o8JoVzW

最后，附上代码,希望大家学习愉快

#include<cstdio>#include<cstring>const int INF=0x3f3f3f3f;int dp[1005][1005];int s[1005][1005];int x[1005];int y[1005];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1; i<=n; i++)            scanf("%d%d",&x[i],&y[i]);        for(int i=1; i<=n; i++)            s[i][i]=i;        for(int len=2; len<=n; len++)            for(int i=1; i<=n-len+1; i++)            {                int j=i+len-1;                dp[i][j]=INF;                for(int k=s[i][j-1];k<=s[i+1][j];k++)                {                    int ans=dp[i][k-1]+dp[k][j]+x[k]-x[i]+y[k-1]-y[j];                    if(dp[i][j]>ans)                    {                        dp[i][j]=ans;                        s[i][j]=k;                    }                }            }        printf("%d\n",dp[1][n]);    }    return 0;}