求矩阵左上角到右下角所经过点的和的最大距离和路径

题目：求一个矩阵的任意两点所经过点的和的最大距离和路径。只能向右或者向下

第一种解法：求出最大距离或者最小距离和经过的路径。(利用动态规划)

import java.util.*;public class Search_T {public static LinkedList<String> list = new LinkedList<String>();public static void main(String[] args){int[][] a = {{1,2,9},    {4,5,9}, {7,8,9}};System.out.println(search(a));}public static int search(int[][] a){String[][] str = new String[a.length][a.length];//String str=""+a[0][0];int[][] k = new int[a.length][a.length];for(int i=0;i<k.length;i++)for(int j=0;j<k.length;j++){if(i==0&&j==0){k[i][j]=1;str[i][j]=""+k[i][j];continue;}if(i==0){k[i][j]=a[i][j]+k[0][j-1];str[i][j]=str[0][j-1]+"->"+a[i][j];continue;}if(j==0){k[i][j]=a[i][j]+k[i-1][0];str[i][j]=str[i-1][0]+"->"+a[i][j];continue;}k[i][j]=Math.max(a[i][j]+k[i-1][j], a[i][j]+k[i][j-1]);if(a[i][j]+k[i-1][j]>a[i][j]+k[i][j-1])str[i][j]=str[i-1][j]+"->"+a[i][j];elsestr[i][j]=str[i][j-1]+"->"+a[i][j];}//for(int i=0;i<k.length;i++)//{//for(int j=0;j<k.length;j++)//{//System.out.printf("%5d",k[i][j]);//}//System.out.println();//}System.out.println(str[2][2]);return k[a.length-1][a.length-1];}}

结果如下：

1->2->9->9->9
30

第二种解法：都可以求出来。（利用图进行搜索，很复杂，但也是求图中两点之间所有路径的通用方法）

import java.util.*;public class Route {private static int[][] graph;private static boolean[] hasFlag;    private static ArrayList<String> res=new ArrayList<String>();public static void main(String[] args){int[][] m={{5 ,14, 3, 4, 5},   {6, 7, 8, 100, 100},   {11,25,16,14,15},   {16,17,18,260,20}};output(m,0,0,3,4);}public static int change(int[][] m,int x,int y){return x*m[0].length+y+1;}public static void output(int[][] m,int xs,int ys,int xd,int yd) //求矩阵m中，从点（xs,ys）到点（xd,yd）经过的点和的最大值。{int sum=0,clu=0,x=0;graph = new int[m.length*m[0].length+1][m.length*m[0].length+1];hasFlag=new boolean[graph.length];for(int i=1;i<graph.length;i++)for(int j=0;j<graph[0].length;j++){graph[i][j]=-1;}for(int i=1;i<m.length*m[0].length;i++){if(i%5==0){graph[i][i+5]=m[solve(i+5,m)[0]][solve(i+5,m)[1]];}else if(i/m.length>=m[0].length-1){graph[i][i+1]=m[solve(i+1,m)[0]][solve(i+1,m)[1]];}else{graph[i][i+1]=m[solve(i+1,m)[0]][solve(i+1,m)[1]];graph[i][i+5]=m[solve(i+5,m)[0]][solve(i+5,m)[1]];}}String str=String.valueOf(change(m,xs,ys));getPaths(change(m,xs,ys),change(m,xd,yd),str,sum);String[] k = new String[res.size()];for(String e:res){k[clu]=e;clu++;}if(res.size()==0){System.out.println("null");return;}      x=judge(k);System.out.println("max="+(Integer.valueOf((k[x].split(":"))[1])+m[xs][ys]));String str1 = (k[x].split(":"))[0];String[] str2=str1.split("->");for(int i=0;i<str2.length;i++){if(i==str2.length-1)System.out.print(m[solve(Integer.parseInt(str2[i]),m)[0]][solve(Integer.parseInt(str2[i]),m)[1]]);elseSystem.out.print(m[solve(Integer.parseInt(str2[i]),m)[0]][solve(Integer.parseInt(str2[i]),m)[1]]+"->");}}public static int judge(String[] a) //选择经过点和最大的一条路径{int max=0;int j=0;for(int i=0;i<a.length;i++){if(Integer.valueOf((a[i].split(":"))[1])>max){max = Integer.valueOf((a[i].split(":"))[1]);j=i;}}return j;}public static void getPaths(int s,int d,String path,int sum) //遍历所有凑够s到d的路径，暴力搜索，穷举法，复杂度很高{hasFlag[s]=true;for(int i=1;i<graph.length;i++){if (graph[s][i]==-1 || hasFlag[i]){continue;}if(i==d){ res.add(path+"->"+d+":"+(sum+graph[s][i]));continue;}getPaths(i, d, path+"->"+i, sum+graph[s][i]);hasFlag[i]=false;}}public static int[] solve(int n,int[][] a)  //讲一个数N转换成坐标x,y{int[] w =new int[2];int x=0,y=0;if(n%a[0].length==0)x=n/a[0].length-1;elsex=n/a[0].length;if(n%a[0].length==0){y=a[0].length-1;}else{y=n%a[0].length-1;}w[0]=x;w[1]=y;return w;}}

结果如下：

max=428
5->14->7->8->100->14->260->20