Introduction to algorithms the 15th chapter of Exercise 15-1--the problem of traveling quotient of two-tune Euclid

Idea: 1), first of all points plus coordinates, x-axis pointing right, y-axis pointing down. All points are then arranged from small to large in x-axis coordinates.

2) The general idea is to remove a node from a well-ordered node in turn, and decide whether the node should be placed on the first path or the second path. 3) define an array: Double b[8][8]; B[I][J] indicates that the first path searches to the node I, and the second path searches to the shortest path length after the J node. If i==j indicates that two paths converge to I point

If the i==n indicates the search to the end

3) method of finding two-point distance: Double Length (Node x[],int i,int J)

4) M[i][j] The shortest distance between the numbered I point and the number J point is stored. Need to declare is: B[i][j]=b[j][i];

So, the M-matrix is a symmetric matrix. That is, the distance between point I and J is equal to the matrix of J Point and I point. So here we just need to ask for the lower triangular matrix B.

5) Recursive formula from the above ideas: (I>j for the lower triangle)

(I==J): B[i][j]=b[i][j-1]+length[i][j-1]
(i>j+1): b[i][j]= B[i-1][j]+length[i-1][i]
(i=j+1): B[i][j]=min (1<=k<j) (B[k][j]+length[k][i])//j==1 b[i][j]=length (I,J);

From the above analysis can get the following code:

Double Tune Euclid travel quotient problem #include <iostream> #include <math.h> #define M 65536 using namespace std;
	Define node coordinates struct node {int x;
int y;
}N[10];
	Find the length of the node I and J of the node double (node *n,int i,int j) {double L;
	L=sqrt (double ((n[i].x-n[j].x) * (n[i].x-n[j].x) + (N[I].Y-N[J].Y) * (N[I].Y-N[J].Y));
return L;
	} void ShortPath (Node*n, double (*b) [10],int length) {int i,j,k;
	Double num;
	Defines the starting node ordinal of 1 b[1][1]=0;
			for (i=2;i<=length;i++) {for (j=1;j<=i;j++) {//If the end of the two path is I, the length of the total path is one from 1 to I and//a path from 1 to I-1 plus the distance from i-1 to I.
			if (i==j) {b[i][j]=b[i][j-1]+length (n,i,j-1); }//If there is more than one point between I and J, then the path of J points will not change, and the path of I point is//path from 1 to I-1 plus i-1 to I path if (i>j+1) {b[i][j]=b[i-1][j]+length (n,i-1,i)
			;
				} if (i==j+1) {b[i][j]=m;
				if (j==1) {b[i][j]=length (n,i,j);
					} for (k=1;k<j;k++) {num=b[k][j]+length (n,k,i);
					if (b[i][j]>num) {b[i][j]=num;
		}}} B[j][i]=b[i][j];
	}}} int main () {n[1].x=0;
	n[1].y=0;N[2].x=1;
	n[2].y=6;
	n[3].x=2;
	n[3].y=3;
	n[4].x=5;
	n[4].y=2;
	n[5].x=6;
	n[5].y=5;
	n[6].x=7;
	N[6].y=1;
	n[7].x=8;
	n[7].y=4;
	Double b[10][10]={0};
	ShortPath (n,b,7);
	cout<<b[7][7]<<endl;
return 0; }