HDOJ 3516 Tree Construction

四边形优化DP

Tree Construction

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 868    Accepted Submission(s): 470

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

 

Source

2010 ACM-ICPC Multi-University Training Contest（8）——Host by ECNU

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1100;const int INF=0x3f3f3f3f;struct POINT{int x,y;}pt[maxn];int n;int dp[maxn][maxn],s[maxn][maxn];inline int cost(int i,int j,int k){return pt[k].y-pt[j].y+pt[k+1].x-pt[i].x;}int main(){while(scanf("%d",&n)!=EOF){for(int i=1;i<=n;i++){scanf("%d%d",&pt[i].x,&pt[i].y);}for(int i=1;i<=n;i++){s[i][i]=i;}for(int len=2;len<=n;len++){for(int i=1;i+len-1<=n;i++){int j=i+len-1;dp[i][j]=INF;for(int k=s[i][j-1];k<=s[i+1][j]&&k<j;k++){if(dp[i][j]>dp[i][k]+dp[k+1][j]+cost(i,j,k)){s[i][j]=k;dp[i][j]=dp[i][k]+dp[k+1][j]+cost(i,j,k);}}}}printf("%d\n",dp[1][n]);}return 0;}