其实它的代码理解起来真的挺难的我觉得!!!
昨天看了一下午感觉晦涩难懂,还是matlab好用,直接调用函数就可以了!!!
不过这里还是得跟大家介绍一下:
1.问题的理解:
像这种带权的有向图,每一行都表示该行标号对应列标号的有向权值,本身到本身的数值为0,没办法到达的数值为∞
2.最优子结构
i,j是图里面的两个不同顶点,设p为从i到j的不经过{k+1,k+2,...n}点的最短路径
(1)若p不经过顶点k,则p也是从i到达j其间不经过{k,k+1,k+2...n}的最短路径
(2)若p经过顶点k,我们把前半段从i到k记为p1,后半段从k到j记做p2,则p1是从i到k不经过{k,k+1,...n}的最短路径,p2是从k到j的其间不经过{k,k+1,k+2...n}的最短路径
算法伪代码:
FLOYD-WARSHALL(w)
1. for i<-1 to n
2. do for j<-1 to n
3. do if i=j or w[i,j]=∞
4. then π[i,j]<-NIL
5. else π[i,j]<-i
6. D<-w
7. for k<-1 to n
8. do for i<- 1 to n
9. do for j<-1 to n
10. if D[i,j]>D[i,k]+D[k,j]
11. then D[i,j]<-D[i,k]+D[k,j]
12. π[i,j]<-π[k,j]
13. return D,π
(吐槽一下,这里的D其实就是每行标号到每列标号的最短距离,后期程序出来后还是很漂亮的,然后这里的π我就不知道什么鬼了,索性就不管他啦,哈哈!!)
PRINT-ALL-PAIRS-SHORTEST -PATHS(π,i,j)
1. if i=j
2. then print i
3. else if π[i,j]=NIL
4. then print"no path from " i “to” j "exists"
5. else PRINT-ALL-PAIRS-SHORTEST-PATHS(π,i,π[i,j])
6. print j
C++:
floyd.h
#define _floyd_h
#include<string.h>
#include<float.h>
#include<stdio.h>
#include"pair.h"
#include<malloc.h>
pair floydwarshall(double* w,int n){
double *d=(double*)malloc(n*n*sizeof(double));
int i,j,k,*pi=(int*)malloc(n*n*sizeof(int));
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(j==i||w[i*n+j]>=DBL_MAX)
pi[i*n+j]=-1;
else
pi[i*n+j]=i;
memcpy(d,w,n*n*sizeof(double));
for(k=1;k<=n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(d[i*n+j]>d[i*n+k-1]+d[(k-1)*n+j]){
d[i*n+j]=d[i*n+k-1]+d[(k-1)*n+j];
pi[i*n+j]=pi[(k-1)*n+j];
}
return make_pair(d,pi);
}
void printallpairsshotestpath(int *pi,int n,int i,int j){
if(i==j){
printf("%d ",i+1);
return;
}
if(pi[i*n+j]==-1)
printf("no path from %d to %d exists.\n",i+1,j+1);
else{
printallpairsshotestpath(pi,n,i,pi[i*n+j]);
printf("%d ",j+1);
}
}
main.cpp
#include<stdlib.h>
#include"floyd.h"
int main(){
double w[]={0,3,8,DBL_MAX,-4,
DBL_MAX,0,DBL_MAX,1,7,
DBL_MAX,4,0,DBL_MAX,DBL_MAX,
2,DBL_MAX,-5,0,DBL_MAX,
DBL_MAX,DBL_MAX,DBL_MAX,6,0},*d;
int i,j,k,*pi,n=5;
pair r;
r=floydwarshall(w,n);
d=(double*)r.first;
pi=(int*)r.second;
for(i=0;i<n;i++){
for(j=0;j<n;j++){
printallpairsshotestpath(pi,n,i,j);
printf(":%1.1f\n",d[i*n+j]);
}
}
free(d);
free(pi);
}
输出结果:
JAVA:
Floyd.java
package Jamin;
import java.util.Arrays;
public class Floyd {
public static Pair floydwarshall(double[][] w) {
int n=w.length,i,j,k;
double[][] d=new double[n][n];
int[][] pi=new int[n][n];
d=Arrays.copyOf(w,n);
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(j==i||w[i][j]>=Double.MAX_VALUE)
pi[i][j]=-1;
else
pi[i][j]=i;
for(k=1;k<=n;k++)
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(d[i][j]>d[i][k-1]+d[k-1][j]) {
d[i][j]=d[i][k-1]+d[k-1][j];
pi[i][j]=pi[k-1][j];
}
return Pair.make(d, pi);
}
public static void printallpairsshotestpath(int[][] pi,int i,int j) {
int n=pi.length;
if(i==j) {
System.out.print((i+1)+" ");
return;
}
if(pi[i][j]==-1)
System.out.println("no path from " +(i+1)+"to "+(j+1)+"exists.");
else {
printallpairsshotestpath(pi,i,pi[i][j]);
System.out.print((j+1)+" ");
}
}
}
Test.java
package Jamin;
public class Test {
public static void main(String[] args) {
// TODO Auto-generated method stub
double[][] w= {{0.0,3.0,8.0,Double.MAX_VALUE,-4.0},
{Double.MAX_VALUE,0.0,Double.MAX_VALUE,1.0,7.0},
{Double.MAX_VALUE,4.0,0.0,Double.MAX_VALUE,Double.MAX_VALUE},
{2.0,Double.MAX_VALUE,-5.0,0.0,Double.MAX_VALUE},
{Double.MAX_VALUE,Double.MAX_VALUE,Double.MAX_VALUE,6.0,0.0}
},d;
int n=5,i,j,k;
int[][] pi;
Pair r;
r=Floyd.floydwarshall(w);
d=(double[][])r.first;
pi=(int[][])r.second;
for(i=0;i<n;i++) {
for(j=0;j<n;j++) {
Floyd.printallpairsshotestpath(pi, i, j);
System.out.println(":"+d[i][j]);
}
}
}
}
输出结果跟上面一样的!
JAVA:(贪心算法版)
探讨本部分时可以参考下该链接:
https://blog.csdn.net/weixin_39653545/article/details/82925768
贪心选择:
由图中内容可知,其贪心选择方式跟最小生成树是一样的:
从第一个起始点开始(生成树的根),遍历剩下的位置点,找到在根附近的点中距离最小的位置点,记录下来放在位置点集合U中,然后在剩下的点中遍历找到位置点集合U中所有位置点的附近点,找到距离最短的,记录下来放在集合U中,重复多次直至所有的点都在集合U中。
Shortestpath.java
package Jamin;
import java.util.Arrays;
public class Shortestpath {
public static Pair dijkstra(double[][] w,int s) {
int n=w.length,i;
Vertex[] d=new Vertex[n];
int[] pi=new int[n];
boolean[] poped=new boolean[n];
Arrays.fill(pi,-1);
Arrays.fill(poped,false);
for(i=0;i<n;i++)
d[i]=new Vertex(Double.POSITIVE_INFINITY,i);
d[s].weight=0.0;
Prioqueue Q=new Prioqueue(new Less());
for(i=0;i<n;i++)
Q.enqueue(d[i]);
while(!Q.empty()) {
int u=((Vertex)(Q.dequeue())).index;
poped[u]=true;
for(int v=0;v<n;v++) {
if(w[u][v]>0.0)
if(!poped[v])
if(d[v].weight>d[u].weight+w[u][v]) {
pi[v]=u;
d[v].weight=d[u].weight+w[u][v];
}
}
Q.fix();
}
return Pair.make(d, pi);
}
public static void printpath(int[] pi,int s,int i) {
if(i==s) {
System.out.print((i+1)+" ");
return;
}
if(pi[i]==-1)
System.out.println("no path form "+(s+1)+"to"+(i+1));
else {
printpath(pi,s,pi[i]);
System.out.print((i+1)+" ");
}
}
}
Test.java
package Jamin;
public class Test {
public static void main(String[] args) {
// TODO Auto-generated method stub
Pair r=new Pair();
double a[][]= {{0,10,0,5,0},
{0,0,1,2,0},
{0,0,0,0,4},
{0,3,9,0,2},
{7,0,6,0,0}};//这里的矩阵表示:第一个为标号1到标号1的距离,第一行第二个为标号1到标号2的距离。。。
int i,n=5,s=0;//这里的n表示有5个标号位置
int[] pi;
Vertex[] d;
r=Shortestpath.dijkstra(a, s);
d=(Vertex[])r.first;
pi=(int[])r.second;
for(i=0;i<n;i++) {
if(i!=s) {
Shortestpath.printpath(pi, s, i);
System.out.println();
System.out.println(d[i].weight);
}
}
}
}
输出结果:
1 4 2 //行走路径
8.0 //总距离
1 4 2 3
9.0
1 4
5.0
1 4 5
7.0