一、問題描述
有一個 n * n 的矩陣，其中有四個2代表研究院，矩陣中 0 表示道路，1 表示不能走，只可以四個方向走，求找到一點（0點）距離四個研究院中最遠的研究院的距離最近。

測試資料：
5
8 8
4
2 0 0 0 0 0 2 0
0 1 0 0 0 0 0 0
0 0 1 1 0 0 0 0
0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0
0 1 1 1 1 1 0 0
0 2 0 0 0 0 0 2
3 3
2
0 1 1
2 1 0
0 0 2
4 4
3
2 0 1 1
0 0 0 0
0 0 1 0
1 2 0 2
10 10
4
0 0 1 0 2 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0
2 0 0 0 1 0 1 0 0 1
0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 1 0 0 1 0
0 1 0 0 1 0 2 0 1 0
0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 0 1 0 0 1
0 1 0 0 0 0 0 0 0 0
1 2 0 0 0 0 0 0 0 0
20 20
4
2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0  
1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 

二、DFS 版本，必定超時
1.思路：
（1）遍歷所有的 0 點，找到距離每一個2的最近距離，儲存下來。
這裡就比較坑了，必須一個2一個2的找，當找任意一個2時，將其他的2設為0，因為2也可以走。這樣相當於每一個0就要單純的找4個2的最近距離，這樣做是比較慢的。
（2）每走一個0點，求出其最遠距離max。
（3）比較所有的0點的最遠距離max，即可求出最終答案。

2.程式碼

package com.samsung.minroads;

import java.io.*;
import java.util.*;
public class MinRoads {

    static
 
 int totalRow, totalCol, totalNum, max, minLen;
    static int[][] data;

    static int[] steps;

    static int[][] dir = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } }; // 右，下，左，上

    static boolean isEnd;

    static Kuang[] kuangList;
    static int kuIn;

    public static void main(String[] args) throws Exception {
        Scanner sc = new
 
 Scanner(new File("src/minroads.txt"));
        int exit = sc.nextInt();
        while ((exit--) >= 0) {
            totalRow = sc.nextInt();
            totalCol = sc.nextInt();
            totalNum = sc.nextInt();
            // init
            data = new int[totalRow][totalCol];
            steps = new int[(totalNum + 1)];
            kuangList = new Kuang[totalNum + 1];
            isEnd = false;
            max = 0;
            minLen = 1000;
            kuIn = 0;

            for (int i = 0; i <= totalNum; i++) {
                steps[i] = 1000;
            }

            for (int i = 0; i < totalRow; i++) {
                for (int j = 0; j < totalCol; j++) {
                    data[i][j] = sc.nextInt();
                    if (data[i][j] == 2) {
                        Kuang ku = new Kuang();
                        ku.x = i;
                        ku.y = j;
                        kuangList[kuIn++] = ku;
                    }
                }
            }

            for (int i = 0; i < totalRow; i++) {
                for (int j = 0; j < totalCol; j++) {
                    if (data[i][j] == 0) {
                        // init
                        // System.out.println("i:" + i + ",j:" + j);
                        for (int x = 0; x <= totalNum; x++) {
                            steps[x] = 1000;
                        }
                        max = 0;
                        for (int curK = 0; curK < kuIn; curK++) {
                            bfs(i, j, 0, 1, curK);
                        }
                        for (int k = 0; k < kuIn; k++) {
                            max = steps[k] > max ? steps[k] : max;
                            //System.out.println(steps[k]);
                        }
                        // System.out.println("max:" + max);
                        minLen = max < minLen ? max : minLen;
                    }
                }
            }
            System.out.println("minLen:" + minLen);
        }
    }

    // 到一個點距離最近
    static void bfs(int row, int col, int curVal, int step, int curK) {
        // System.out.println(step);

        if(step > minLen){    // 剪枝
            return;
        }

        int newRow, newCol;
        for (int i = 0; i < 4; i++) {
            newRow = row + dir[i][0];
            newCol = col + dir[i][1];
            if (newRow >= 0 && newRow < totalRow && newCol >= 0
                    && newCol < totalCol) {
                if (data[newRow][newCol] == 0 || data[newRow][newCol] == 2) {
                    if (data[newRow][newCol] == 2
                            && (newRow == kuangList[curK].x && newCol == kuangList[curK].y)) {
                        steps[curK] = step < steps[curK] ? step : steps[curK];
                        return;
                    }
                    curVal = data[newRow][newCol];
                    data[newRow][newCol] = 8; // 8 不能走
                    bfs(newRow, newCol, curVal, step + 1, curK);
                    data[newRow][newCol] = curVal;
                }
            }
        }
    }

    static void print() {
        for (int i = 0; i < totalRow; i++) {
            for (int j = 0; j < totalCol; j++) {
                System.out.print(data[i][j] + " ");
            }
            System.out.println();
        }
        System.out.println();
    }
}

class Kuang {
    int x;
    int y;
}

三、BFS 版本
方法一：
1.思路：
（1）遍歷所有的0點，記錄每個元素被遍歷的步數，所以點走完一個0點時就可以得到4個2分別的步數，比較這4個步數獲取最遠距離max即可。
（2）遍歷完所有的0點，就可以獲取到最終的答案。

2.程式碼

package com.samsung.minroads;

import java.io.File;
import java.util.Scanner;

public class MinRoads2 {

    static int totalRow, totalCol, totalNum,max,minLen;
    static int[][] data,newData;

    static int[][] dir = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } }; // 右，下，左，上

    static Stone[] queue = new Stone[500];   // 走過的點
    static int start,end;

    static Stone[] stoneList;   // 研究中心
    static int stIn;

    public static void main(String[] args) throws Exception{
        Scanner sc = new Scanner(new File("src/roads.txt"));
        int exit = sc.nextInt();
        while ((exit--) != 0) {
            totalRow = sc.nextInt();
            totalCol = sc.nextInt();
            totalNum = sc.nextInt();
            // init
            data = new int[totalRow][totalCol];
            newData = new int[totalRow][totalCol];
            stoneList = new Stone[totalNum + 1];
            stIn = 0;
            max = 0;
            minLen = 1000;

            for (int i = 0; i < totalRow; i++) {
                for (int j = 0; j < totalCol; j++) {
                    data[i][j] = sc.nextInt();
                    newData[i][j] = data[i][j];
                    if (data[i][j] == 2) {  
                        Stone stone = new Stone();
                        stone.x = i;
                        stone.y = j;
                        stone.size = 1000;   // 方便計算
                        stoneList[stIn++] = stone;
                    }
                }
            }

            for (int i = 0; i < totalRow; i++) {
                for (int j = 0; j < totalCol; j++) {
                    if(data[i][j] == 0){
                        //System.out.println("i" + i + ",j:" + j);
                        for(int x = 0;x < 500;x++){
                            queue[x] = null;
                        }
                        start = 0;
                        end = 1;
                        max = 0;
                        Stone startSt = new Stone();
                        startSt.x = i;
                        startSt.y = j;
                        startSt.size = 1;
                        queue[0] = startSt;
                        bfs(0,1);

                        for(int k = 0;k < totalNum;k++){
                            Stone st = stoneList[k];
                            if(null != st){
                                max = st.size > max ? st.size : max;
                                st.size = 1000;
                            }
                        }
                        minLen = max < minLen ? max : minLen;
                        for (int m = 0; m < totalRow; m++) {
                            for (int n = 0; n < totalCol; n++) {
                                data[m][n] = newData[m][n];
                            }
                        }
                    }
                }
            }
            System.out.println("minLen:" + minLen);
        }
    }

    static void bfs(int step,int curStep){
        int newRow, newCol;
        int temp = 0;
        Stone sto = queue[start++];
        if(null != sto){
            temp = sto.size;
            if(temp >= step){
                step = step + 1;
            }

            for (int i = 0; i < 4; i++) {
                newRow = sto.x + dir[i][0];
                newCol = sto.y + dir[i][1];
                if (newRow >= 0 && newRow < totalRow && newCol >= 0
                        && newCol < totalCol) {
                    if(data[newRow][newCol] != 1){

                        for(int k = 0;k < stIn;k++){
                            Stone exiStone = stoneList[k];
                            if(newRow == exiStone.x && newCol == exiStone.y){
                                exiStone.size = step < exiStone.size ? step : exiStone.size;
                                stoneList[k] = exiStone;
                            }
                        }
                        Stone st = new Stone();
                        st.x = newRow;
                        st.y = newCol;
                        st.size = step;
                        queue[end++] = st;
                        data[newRow][newCol] = 1;
                    }
                }
            }
            bfs(step,sto.size);
        }       
    }
}

class Stone{
    int x;
    int y;
    int size;
}

方法二
1.思路
從2開始遍歷，記錄每個2到所有0點的距離，也是採用bfs遍歷。這樣做可以少走很多重複的路，複雜度和時間度都相對不負責。