钓鱼 - 算法（暂存）

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1. 钓鱼题
   总共N个座位，三个门，每个门后面有M个人。进入的规则：门的顺序可以变，但是每个门打开后，门后的人必须全部进去才能开下一个门。每个人只能坐在离他最近的座位上，求人全部进入所走的最小步数

5 –test case
10 –总共的座位数
4 5 – 第一个门的位置是4，有5个人
6 2 – 第二个门的位置是6，有2个人
10 2 – 地三个门的位置是10，有2个人
10 – 下一组case
8 5
9 1
10 2
24
15 3
20 4
23 7
39
17 8
30 5
31 9
60
57 12
31 19
38 16

1 18

2 25

3 57

4 86

5 339

package com.daxiong.fish;

import java.io.*;
import java.util.*;

public class Fish {

    static int[] data;
    static int sit;
    static int[] door;
    static int minSteps;

    static int[] orgin;
    static int[] allPer;
    static int[] used;

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(new File("src/case/fish.txt"));
        int door_one,door_sec,door_thr;
        int exit = sc.nextInt();
        while ((exit--) != 0) {
            sit = sc.nextInt();
            data = new int[sit + 10];
            door = new int[sit + 10];
            door_one = sc.nextInt();
            door[door_one] = sc.nextInt();
            door_sec = sc.nextInt();
            door[door_sec] = sc.nextInt();
            door_thr = sc.nextInt();
            door[door_thr] = sc.nextInt();

            minSteps = 1000;
            allPer = new int[100];
            used = new int[3];
            orgin = new int[3];

            orgin[0] = door_one;
            orgin[1] = door_sec;
            orgin[2] = door_thr;

            allPer(0);
            System.out.println(".....");
        }
    }

    // 1 2 3 4 5 6 7 8 9 10
    //       5   2       2
    // 1 2 3 2 3 2 3 1 2 = 
    static void bfs(int[] curData, int doorIndex, int curIndex, int num,int sumStep) {

        System.out.println(doorIndex + "," + sumStep);

        if (num == 0) { // 下一个门
            if(door[curData[doorIndex+1]] > 0){
                bfs(curData, doorIndex + 1, curIndex, door[curData[doorIndex+1]],sumStep);
            }
            if (doorIndex == 2) { // 3 个门 结束
                minSteps = sumStep < minSteps ? sumStep : minSteps;
            }
            return;
        }

        int i = 0,j = 0,before = 0,after = 0;
        boolean isLeft = false,isRight = false;
        for (i = curData[doorIndex]; i > 0; i--) { // 遍历左边
            if (data[i] != 1) {
                before = curData[doorIndex] - i;
                isLeft = true;
                break;
            }
        }
        for (j = curData[doorIndex]; j <= sit; j++) { // 右边
            if (data[j] != 1) {
                after = j - curData[doorIndex];
                isRight = true;
                break;
            }
        }
        // 如果左边或者右边没有了
        if(!isLeft){  // 左边没有,放右边
            data[j] = 1;
            bfs(curData, doorIndex, curIndex, num - 1,sumStep + after + 1);
        } 
        if(!isRight){  // 右边没有了，放左边
            data[i] = 1;
            bfs(curData, doorIndex, curIndex, num - 1,sumStep + before + 1);
        }
        if (after >= 0 && after <= sit && before >= 0 && before <= sit) {
            // 判断距离
            if ((before == after)) {
                if (after == 0) { // 当前点
                    data[i] = 1;
                    bfs(curData, doorIndex, curIndex, num - 1,sumStep + 1);
                } else { // 不同的两点
                    if(num-2 >= 0){
                        data[i] = data[j] = 1;
                        bfs(curData, doorIndex, curIndex, num - 2,sumStep + before + after + 2);
                    } else {
                        data[i] = 1;  // 先放左边
                        bfs(curData, doorIndex, curIndex, num - 1,sumStep + before + 1);
                    }
                }
            } else if (before < after) { // 放到左边
                data[i] = 1;
                bfs(curData, doorIndex, curIndex, num - 1,sumStep + before + 1);
            } else { // 右边
                data[j] = 1;
                bfs(curData, doorIndex, curIndex, num - 1,sumStep + after + 1);
            }
        }
    }

    // 全排列
    static void allPer(int index){
        if(index == 3){
            minSteps = 1000;
            for(int i = 0;i < (sit + 10);i++){
                data[i] = 0;
            }
            bfs(allPer, 0, 0, door[allPer[0]],0);
            System.out.println(minSteps);
            return;
        }
        for(int i = 0;i < 3;i++){
            if(used[i] != 1){
                used[i] = 1;
                allPer[index] = orgin[i];
                allPer(index+1);
                used[i] = 0;
            }
        }
    }

    static void print() {
        System.out.println("--------------");
        for (int i = 1; i <= sit; i++) {
            System.out.print(data[i] + " ");
        }
        System.out.println();
        System.out.println("--------------");
    }

}