2014 WAP校园招聘笔试题1

Abstract
We are planning an orienteering game.
The aim of this game is to arrive at the goal (G) from the start (S) with the shortest distance.
However, the players have to pass all the checkpoints (@) on the map.
An orienteering map is to be given in the following format.
########
#@....G#
##.##@##
#..@..S#
#@.....#
########
In this problem, an orienteering map is to be given.
Calculate the minimum distance from the start to the goal with passing all the checkpoints.
Specification
* A map consists of 5 characters as following.
You can assume that the map does not contain any invalid characters and
the map has exactly one start symbol 'S' and exactly one goal symbol 'G'.
* 'S' means the orienteering start.
* 'G' means the orienteering goal.
* '@' means an orienteering checkpoint.
* '.' means an opened-block that players can pass.
* '#' means a closed-block that players cannot pass.
* It is allowed to move only by one step vertically or horizontally (up, down, left, or right) to the
next block.
Other types of movements, such as moving diagonally (left up, right up, left down and right down)
and skipping one or more blocks, are NOT permitted.
* You MUST NOT get out of the map.
* Distance is to be defined as the number of movements to the different blocks.
* You CAN pass opened-blocks, checkpoints, the start, and the goal more than once if necessary.
* You can assume that parameters satisfy following conditions.
* 1 <= width <= 100
* 1 <= height <= 100
* The maximum number of checkpoints is 18.
* Return -1 if given arguments do not satisfy specifications, or players cannot arrive at the goal

from the start by passing all the checkpoints.

import java.util.HashMap;import java.util.LinkedList;import java.util.Map;import java.util.Scanner;class node {int x, y;int index;int step;};/** * 状态压缩DP，总共的状态数量为2^（checkPointsNum+1）个 *  * @created Oct 29, 2014 * @author  chenshanfu */public class Orienteering_ori {private static int height, width;private static int[][] shortestDistance;// checkPoints之间的最短距离private static int[][] f;// f[k][state] 到达第k个点的所经过的状态为state的private static int checkPointsNum;private static int stateNum;private static boolean[][] isvisited;//记录isvisited[i][j]是否已经遍历过// 增加一层映射关系，@坐标-->index 映射private static Map<Integer, Integer> map = new HashMap<Integer, Integer>();private static node startPoint = new node();private static node endPoint = new node();private static char[][] a;// 原始地图private static int[][] dir = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };//可以遍历的4个方向private static boolean isException;//输入的参数异常private static int startPointsNum = 0;private static int endPointsNum = 0;/** * 读取数据，同时进行参数初始化 *  * @created Oct 29, 2014 * @author  chenshanfu */@SuppressWarnings("resource")public static void init() {Scanner input = new Scanner(System.in);String line = "";line = input.nextLine().trim();if (line.split(" ").length != 2) {isException = true;}else {width = Integer.parseInt(line.split(" ")[0]);height = Integer.parseInt(line.split(" ")[1]);if (width < 1 || width > 100 || height < 1 || width > 100) {isException = true;}a = new char[height][width];int checkPointsIndex = 0;for (int i = 0; i < height; i++) {line = input.nextLine().trim();if (width != line.length()) {isException = true;break;}for (int j = 0; j < width; j++) {char c = line.charAt(j);a[i][j] = c;int tmp = i * width + j;if (c == '@') {map.put(tmp, ++checkPointsIndex);} else if (c == 'S') {startPoint.x = i;startPoint.y = j;startPoint.step = 0;startPoint.index = tmp;startPointsNum++;} else if (c == 'G') {endPoint.x = i;endPoint.y = j;endPoint.step = 0;endPoint.index = tmp;endPointsNum++;}}}checkPointsNum = checkPointsIndex;if (startPointsNum != 1 || endPointsNum != 1 || checkPointsNum < 0|| checkPointsNum > 18) {isException = true;}stateNum = (int) Math.pow(2.0, 1.0 * checkPointsNum + 1);map.put(startPoint.index, 0);map.put(endPoint.index, checkPointsNum + 1);/* * System.out.println("stateNum:"+stateNum); * System.out.println("checkPointsNum:"+checkPointsNum); for(Integer * s:map.keySet()){ System.out.println("**"+s+":"+map.get(s)); } */f = new int[checkPointsNum + 2][stateNum];isvisited = new boolean[height][width];shortestDistance = new int[checkPointsNum + 2][checkPointsNum + 2];// shortestDistance// 初始for (int i = 0; i < checkPointsNum + 2; i++) {for (int j = i + 1; j < checkPointsNum + 2; j++) {shortestDistance[i][j] = Integer.MAX_VALUE;shortestDistance[j][i] = Integer.MAX_VALUE;}}}}/** *  * 计算初始节点s到其他checkPoints之间的最短距离，利用BFS实现 * @param s * @created Oct 29, 2014 * @author  chenshanfu */public static void bfs(node s) {LinkedList<node> queue = new LinkedList<node>();queue.add(s);for (int i = 0; i < height; i++) {for (int j = 0; j < width; j++) {isvisited[i][j] = false;}}isvisited[s.x][s.y] = true;while (!queue.isEmpty()) {node front = new node();front = queue.getFirst();queue.removeFirst();for (int i = 0; i < 4; i++) {node tmpnode = new node();tmpnode.x = front.x + dir[i][0];tmpnode.y = front.y + dir[i][1];if (tmpnode.x >= 0 && tmpnode.x < height && tmpnode.y >= 0&& tmpnode.y < width&& !isvisited[tmpnode.x][tmpnode.y]&& a[tmpnode.x][tmpnode.y] != '#') {isvisited[tmpnode.x][tmpnode.y] = true;tmpnode.step = front.step + 1;tmpnode.index = tmpnode.x * width + tmpnode.y;queue.add(tmpnode);if (map.get(tmpnode.index) != null) {shortestDistance[map.get(s.index)][map.get(tmpnode.index)] = tmpnode.step;}}}}}/** * 计算所有的checkPoints之间的最小距离 *  * @created Oct 29, 2014 * @author  chenshanfu */public static void getMinDistance() {for (int i = 0; i < height; i++) {for (int j = 0; j < width; j++) {if (a[i][j] == '@' || a[i][j] == 'S' || a[i][j] == 'G') {node tmpNode = new node();tmpNode.x = i;tmpNode.y = j;tmpNode.index = i * width + j;tmpNode.step = 0;bfs(tmpNode);}}}// 初始化@到S的最短距离for (int i = 1; i < checkPointsNum + 2; i++) {f[i][0] = shortestDistance[i][0];}}/** * 打印最短路径信息 *  * @created Oct 29, 2014 * @author  chenshanfu */public static void print() {for (int i = 0; i < checkPointsNum + 2; i++) {for (int j = 0; j < checkPointsNum + 2; j++) {System.out.print(shortestDistance[i][j] + " ");}System.out.println();}}/** * 状态压缩DP，求解到第k个点并且经过state状态的最优解，同时记录中间递归结果，加速搜索进度 *  * @param k * @param state * @return * @created Oct 29, 2014 * @author  chenshanfu */public static int solve(int k, int state) {// System.out.println("k="+k+" state:"+state);if (f[k][state] > 0 && f[k][state] < Integer.MAX_VALUE)return f[k][state];else {int ans = Integer.MAX_VALUE;for (int i = 1; i < checkPointsNum + 2; i++) {if ((state & (int) Math.pow(2.0, 1.0 * (i - 1))) != 0) {int subsolve = solve(i,state ^ (int) Math.pow(2.0, 1.0 * (i - 1)));//最好用(1 << (i - 1)))位运算来实现，方便快捷if (shortestDistance[i][k] != Integer.MAX_VALUE&& subsolve != Integer.MAX_VALUE) {int sub = subsolve + shortestDistance[i][k];if (sub < ans) {ans = sub;}}}}f[k][state] = ans;return ans;}}public static void main(String[] args) {// TODO Auto-generated method stubinit();if (isException) {System.out.println("-1");} else {getMinDistance();// print();if (isException|| shortestDistance[0][checkPointsNum + 1] == Integer.MAX_VALUE) {System.out.println("-1");} else {int ans = solve(checkPointsNum + 1,(int) Math.pow(2.0, 1.0 * checkPointsNum) - 1);if (ans == Integer.MAX_VALUE)System.out.println("-1");elseSystem.out.println(ans);}}}}

testData：

8 6#########@....G###.##@###..@..S##@.....#########188 6#########@....G##########..@..S##@.....#########-15 4######...##S#G######45 5######.@@##S####..G######95 5######S..###G###..@######65 5####@#.@@##S####..G######-18 6#########@.S..G######@###..##...#@....@#########238 6#########@....G######@###..S#...#@....@#########218 6#########S.S..G######@###..##...#@....@#########-18 6#########@.S..G##########..##...#@....@#########-18 6#########@.S..G.#######.#..##...#@....@#########278 6#########@.S..G.#.#####.#..##...#@....@#########15