Tree Construction - HDU 3516 四边形不等式优化

Tree Construction

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 941 Accepted Submission(s): 510



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

5
1 5
2 4
3 3
4 2
5 1
1
10000 0

Sample Output

12
0

题意：用一棵树把所有的点都连接起来。求最小的所需线段的长度和。

思路：设dp[i][j]为合并从i到j的点所需的最小权值。

dp[i][j]=min(dp[i][k]+dp[k+1][j]+x[k+1]-x[i]+y[k]-y[j])。

然后利用四边形不等式去优化。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t,n,m;
int dp[1010][1010],s[1010][1010],x[1010],y[1010],INF=1e9;
int main()
{
    int i,j,k,ret,len;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
           scanf("%d%d",&x[i],&y[i]);
        for(i=1;i<=n;i++)
           s[i][i]=i-1;
        for(len=1;len<n;len++)
           for(i=1;i+len<=n;i++)
           {
               j=i+len;
               dp[i][j]=INF;
               for(k=s[i][j-1];k<=s[i+1][j];k++)
               {
                   ret=dp[i][k]+dp[k+1][j]+x[k+1]-x[i]+y[k]-y[j];
                   if(ret<dp[i][j])
                   {
                       dp[i][j]=ret;
                       s[i][j]=k;
                   }
               }
           }
        printf("%d\n",dp[1]
);
    }
}