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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.
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
#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; }
HDOJ 3516 Tree Construction
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