dfs
作者：互联网
1.迷宫
public class Point {
int x;
int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
}
public class keda2 {
List<Point> temp =new LinkedList<>();
List<Point> result =new LinkedList<>();
int minCount =Integer.MAX_VALUE;
public ArrayList<Point> winMazeGift(int[][] maze) {
// write code here
Point gift =null;
boolean flag=false;
int n = maze.length;
int m = maze[0].length;
ArrayList<Point> ans = new ArrayList<>();
//找到礼物点
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 4; j++) {
if (maze[i][j] == 8) {
gift=new Point(i,j);
flag=true;
temp.add(new Point(i,j));
break;
}
}
if(flag) break;
}
//从礼物点回溯
dfs(gift,0,maze);
//正序输出
ArrayList<Point> finalResult=new ArrayList<>();
while (!result.isEmpty()){
finalResult.add(result.get(result.size()-1));
result.remove(result.size()-1);
}
return finalResult;
}
private void dfs(Point gift,int count,int [][] maze){
//若走到边界，说明发现了路径，比较是否最短
if(gift.x==0||gift.x==3||gift.y==0||gift.y==3){
if(count< minCount){
result =new LinkedList<>(temp);
minCount =count;
}
return;
}
//判断下个点能不能走，上下左右四个方向，
if(maze[gift.x-1][gift.y]==0){
Point point1=new Point(gift.x-1,gift.y);
//下个点加入路径
temp.add(point1);
//将当前点设置为-1，表示不可走(不加的话来回走，超时)
int kk=maze[gift.x][gift.y];
maze[gift.x][gift.y]=-1;
//向下递归
dfs(point1,count+1,maze);
maze[gift.x][gift.y]=kk;
temp.remove(temp.size()-1);
}
if(maze[gift.x+1][gift.y]==0){
Point point2=new Point(gift.x+1,gift.y);
temp.add(point2);
int kk=maze[gift.x][gift.y];
maze[gift.x][gift.y]=-1;
dfs(point2,count+1,maze);
maze[gift.x][gift.y]=kk;
temp.remove(temp.size()-1);
}
if(maze[gift.x][gift.y-1]==0){
Point point3=new Point(gift.x,gift.y-1);
temp.add(point3);
int kk=maze[gift.x][gift.y];
maze[gift.x][gift.y]=-1;
dfs(point3,count+1,maze);
maze[gift.x][gift.y]=kk;
temp.remove(temp.size()-1);
}
if(maze[gift.x][gift.y+1]==0){
Point point3=new Point(gift.x,gift.y+1);
temp.add(point3);
int kk=maze[gift.x][gift.y];
maze[gift.x][gift.y]=-1;
dfs(point3,count+1,maze);
maze[gift.x][gift.y]=kk;
temp.remove(temp.size()-1);
}
}
}
2.小数
public class keda {
public static void main(String[] args) {
seqSum(2);
}
public static float seqSum (int n) {
double p=1;
double q=2;
double total=0;
double sum=0;
for(int i=1;i<=n;i++){
total=p/q;
double temp=q;
q=p+q;
p=temp;
sum+=total;
}
NumberFormat nf=NumberFormat.getNumberInstance();
nf.setMaximumFractionDigits(2);
// return (float) String.format("%。2f",total);
return Float.parseFloat(nf.format(total));
}
}
3.
两个正整数a，b；有一种操作：删除这两个数中的一个数位，使得a是b的倍数或者b是a的倍数。问最少要操作几次。如 a = 1246，b = 9；可以删除a的一个数位得到126，是9的倍数，因此最少操作一次。 一个字符串，只包含'r'、'e'、'd'三种字符，有一种操作：可以将一个字符替换为其他两种字符。规定字符'e'的左右两个字符有'r'和'd'的，就叫做好'e'。现在需要使字符串中的好'e'数量最多，问最少需要操作多少次？ 举例：reddr，操作一次变成 reder，有两个好'e'。 类似于锯齿数组。给一个数组，每次可以对一个元素值加1操作，现在需要保证数组中的每个元素的左右两个元素相等，且中间的元素不等于左右两个元素。问最少需要操作多少次。 网易给出的样例是： [1,1,4,5,1,4]，最终变成[4,5,4,5,4,5]。为什么到处都有homo啊(恼)。 又是三元组问题。一个数组，n <= 100000，arr[i] <= 10^9；需要找到这样的三元组<i,j,k>，i < j < k，且 ai == ak && ai > aj。问有多少个。
1.
public class aug1 {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int a= in.nextInt();
int b=in.nextInt();
String valuea=a+"";
String valueb=b+"";
int n=valuea.length();
int m=valueb.length();
List<String> ans1= cutDownList(valuea.toString());
List<String> ans2= cutDownList(valueb.toString());
int res=Integer.MAX_VALUE;
for (int i = 0; i <ans1.size() ; i++) {
for (int j = 0; j < ans2.size() ; j++) {
String tempa=ans1.get(i);
String tempb=ans2.get(j);
if(is(Integer.parseInt(tempa),Integer.parseInt(tempb))){
int tmp=n-tempa.length()+m-tempb.length();
res=Math.min(res,tmp);
}
}
}
System.out.println(res);
}
public static boolean is(int a,int b){
if(a%b==0 || b%a ==0) return true;
else return false;
}
public static List<String> cutDownList(String str){
char [] chars=str.toCharArray();
String path="";
List<String> ans=new ArrayList<>();
process(chars,0,ans,path);
return ans;
}
public static void process(char[] chars,int index,List<String> ans,String path){
if(index ==chars.length){
if(path.equals("")) return;
ans.add(path);
return;
}
process(chars,index+1,ans,path);
process(chars,index+1,ans,path+chars[index]);
}
}
2.
/**
* 输入6
* 1 1 4 5 1 4
* 输出11
* {得到长城 {4,5,4,5,4,5}}
*/
public class aug2 {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int nums[]=new int[n];
for (int i = 0; i <n ; i++) {
nums[i]=sc.nextInt();
}
int temp1=0,temp2=0;
//long ans1=0,ans2=0;
for (int i = 0; i <n ; i++) {
if(i%2==1){
//ans1+=nums[i];
temp1=Math.max(temp1,nums[i]);
}
else {
// ans2+=nums[i];
temp2=Math.max(temp2,nums[i]);
}
}
if(temp1==temp2){
temp2++;
}
long res=0;
for (int i = 0; i <n ; i++) {
if(i%2==1){
res+=temp1-nums[i];
}
else {
res+=temp2-nums[i];
}
}
System.out.println(res);
}
}
3.
public class aug3 {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
String str=in.nextLine();
char [] ch=str.toCharArray();
int n=ch.length;
if(n%2==0){
long ans=func(ch,0,n-2,'a');
for (int i = 0; i <n ; i++) {
long tmp=func(ch,0,i,'a')+func(ch,i+1,n-1,'a');
ans=Math.min(ans,tmp);
}
System.out.println(ans);
}else {
long ans=func(ch,0,n-1,'a');
System.out.println(ans);
}
return;
}
public static long func(char[] ch,int begin,int end,char first) {
if(first == 'a'){
return Math.min(func(ch,begin,end,'r'),func(ch,begin,end,'d'));
}
if(begin>=end){
return 0;
}
char next=first;
long cnt =0;
for (int i = begin; i <=end; i++) {
if((i-begin)%2 == 0){
if(ch[i] != next){
cnt++;
}
next =next=='r'?'d':'r';
}else {
if(ch[i] != 'e'){
cnt++;
}
}
}
return cnt;
}
}
4.
public class aug4 {
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
int n=in.nextInt();
int nums[]=new int[n];
for (int i = 0; i < n; i++) {
nums[i]=in.nextInt();
}
int res=0;
for (int i = 0; i <n ; i++) {
int sum=0;
for (int j = i+1; j <n ; j++) {
if(nums[j]<nums[i]){
sum++;
}
else if(nums[j]==nums[i]){
res+=sum;
}
}
}
System.out.println(res);
}
}
5.zijie2
package com.zijie;
import java.util.Scanner;
/**
* @Author:txd
* @Date:2022/8/21
* @Description：
*/
public class aug21_2 {
public static int [][] visited;
public static int x,y;
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int m=sc.nextInt();
char [][] positions=new char[n][m];
int [][] visitor=new int[n][m];
for (int i = 0; i < n ; i++) {
for (int j = 0; j < m ; j++) {
positions[i][j]=sc.next().charAt(0);
}
}
//找到出口点 xStart yStart
for (int i = 0; i < n ; i++) {
for (int j = 0; j < m ; j++) {
if(positions[i][j]=='O'){
y=j;
x=i;
visitor[i][j]=1;
}
}
}
dfs(x-1,y,'D',positions,visitor);
dfs(x+1,y,'U',positions,visitor);
dfs(x,y-1,'R',positions,visitor);
dfs(x,y+1,'L',positions,visitor);
int ans=0;
for (int i = 0; i < n ; i++) {
for (int j = 0; j < m ; j++) {
if(visitor[i][j]==1){
ans++;
}
}
}
System.out.println(n*m-ans);
}
public static void dfs(int i,int j,char c,char position [][],int visitor[][]){
if(i==x&&j==y){
return;
}
if(i<0 || i>=position.length ||j<0 ||j>=position[0].length){
return;
}
if(position[i][j]=='.' || position[i][j]==c){
visitor[i][j]=1;
dfs(i-1,j,'D',position,visitor);
dfs(i+1,j,'U',position,visitor);
dfs(i,j-1,'R',position,visitor);
dfs(i,j+1,'L',position,visitor);
}
}
}
标签：temp,gift,int,dfs,new,maze,public 来源： https://www.cnblogs.com/ttxxdd/p/16611546.html