hdu 3516 Tree construction DP+四边形不等式
太感人了！
hdu 3516，你终于给我过了……
四边形不等式……看了许久，总算用上了一回，这道题和石子合并http://blog.csdn.net/abcjennifer/archive/2010/08/11/5805330.aspx的思路是一模一样的，详解见我上一篇《动态规划经典——石子归并》。
由于点都是从左上往右下来的（Consider a two-dimensional space with a set of points (xi, yi) that satisfy xi < xj and yi > yj for all i < j. ）且要求最后合并成一堆，所以和石子合并一样，每次相邻的两个才合并。
首先最基本的思路是DP，状态转移方程
其中f[i,j]是从i到j合并连续点的最优解，sum[i,j]是合并者相邻两堆需要加上的代价。
现在的复杂度是O(n3)，详细代码见上一篇
[算法优化]：
怎么把复杂度降为O(n2)呢?
优化k！（k是i..j的分割点）
原来我是从i...j-1枚举k, 取f(i,j)=min{f[i,k]+f[k+1,j]}+sum(i,j);
而实际上我只需要从s[i,j-1]到s[i+1,j]枚举k就可以了。其中s[i,j]指f[i...j]取最优解的分割点，这样复杂度就变成O(n2)啦！
Tree Construction
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 273 Accepted Submission(s): 125
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.
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
Source
2010 ACM-ICPC Multi-University Training Contest（8）——Host by ECNU
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zhouzeyong
#include <iostream> using namespace std; #define M 1005 #define INF 1000000000 int n,f[M][M],sum[M][M],s[M][M]; struct POINTS { int x,y; }stone[M]; int dis(POINTS a,POINTS b,POINTS c,POINTS d)//calculate from a to d; b,c is division point { return abs(c.x-a.x)+abs(d.y-b.y); } int main() { int i,j,k,t; while(cin>>n) { for(i=1;i<=n;i++) scanf("%d%d",&stone[i].x,&stone[i].y); memset(f,0,sizeof(f)); for(i=1;i<=n;i++) { s[i][i]=i; } for(int len=2;len<=n;len++)//归并的石子长度 { for(i=1;i<=n-len+1;i++)//i为起点，j为终点 { j=i+len-1; f[i][j]=INF; for(k=s[i][j-1];k<=s[i+1][j];k++) { if(f[i][j]>f[i][k]+f[k+1][j]+dis(stone[i],stone[k],stone[k+1],stone[j])) { f[i][j]=f[i][k]+f[k+1][j]+dis(stone[i],stone[k],stone[k+1],stone[j]); s[i][j]=k; } } } } printf("%d/n",f[1][n]); } return 0; }