路径最优问题

一、问题描述
有一个 n * n 的矩阵，其中有四个2代表研究院，矩阵中 0 表示道路，1 表示不能走，只可以四个方向走，求找到一点（0点）距离四个研究院中最远的研究院的距离最近。

测试数据：58 842 0 0 0 0 0 2 00 1 0 0 0 0 0 00 0 1 1 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 00 0 0 0 0 0 1 00 1 1 1 1 1 0 00 2 0 0 0 0 0 23 320 1 12 1 00 0 24 432 0 1 10 0 0 00 0 1 01 2 0 210 1040 0 1 0 2 0 1 0 0 00 0 0 0 0 0 0 0 0 02 0 0 0 1 0 1 0 0 10 0 0 0 0 0 0 0 0 01 0 0 1 0 1 0 0 1 00 1 0 0 1 0 2 0 1 00 0 0 0 0 0 0 0 0 01 0 0 1 0 0 1 0 0 10 1 0 0 0 0 0 0 0 01 2 0 0 0 0 0 0 0 020 2042 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 20 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 00 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 10 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 01 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 01 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

二、DFS 版本，必定超时
1.思路：
（1）遍历所有的 0 点，找到距离每一个2的最近距离，保存下来。
这里就比较坑了，必须一个2一个2的找，当找任意一个2时，将其他的2设为0，因为2也可以走。这样相当于每一个0就要单纯的找4个2的最近距离，这样做是比较慢的。
（2）每走一个0点，求出其最远距离max。
（3）比较所有的0点的最远距离max，即可求出最终答案。

2.代码

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);        }    }    // 到一个点距离最近    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）遍历所有的0点，记录每个元素被遍历的步数，所以点走完一个0点时就可以得到4个2分别的步数，比较这4个步数获取最远距离max即可。
（2）遍历完所有的0点，就可以获取到最终的答案。

2.代码

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];   // 走过的点    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;   // 方便计算                        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
从2开始遍历，记录每个2到所有0点的距离，也是采用bfs遍历。这样做可以少走很多重复的路，复杂度和时间度都相对不负责。