杭电ACM OJ 1026 Ignatius and the Princess I DFS+BFS
Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20079 Accepted Submission(s): 6540
Special Judge
Problem Description
The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:
1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.
1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.
Input
The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.
Output
For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.
Sample Input
5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX. 5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX1 5 6 .XX... ..XX1. 2...X. ...XX. XXXXX.
Sample Output
It takes 13 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) FINISH It takes 14 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) 14s:FIGHT AT (4,5) FINISH God please help our poor hero.
FINISH
翻译：
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
这样一个迷宫，起点是00 终点是右下角，1,2等数字是怪物。走一步需要1s，杀怪物如果显示的是1需要1秒，如果2需要2秒，以此类推。
思路：
BFS+优先队列剪枝 这里先不剪枝 分别用DFS BFS两个暴力法做 以后再写优先队列剪枝的情况
DFS就是深度优先搜索，把一个方向的所有可能都找完了再找另一个方向。十分简单粗暴。
BFS就是广度优先搜索，所有的点同时前进。这里要注意杀怪时间即可，如果怪物时间没跑完，你是不能继续走的。
DFS代码
void back(int time, int row, int col) {//origin time = 0//row 行 // col 列 int killTime = 1; if (!a[row][col].equals(".")) { killTime += Integer.valueOf(a[row][col]); } if (col == width - 1 && row == height - 1) { min = Math.min(min, killTime + time); return; } for (int i = 0; i < 4; i++) { switch (i) { case 0: int aim = row + 1; if (isAccessable(aim, col)) { isAccessed[aim][col] = true; back(time + killTime, aim, col); isAccessed[aim][col] = false; } break; case 1: int aim2 = row - 1; if (isAccessable(aim2, col)) { isAccessed[aim2][col] = true; back(time + killTime, aim2, col); isAccessed[aim2][col] = false; } break; case 2: int aim3 = col + 1; if (isAccessable(row, aim3)) { isAccessed[row][aim3] = true; back(time + killTime, row, aim3); isAccessed[row][aim3] = false; } break; case 3: int aim4 = col - 1; if (isAccessable(row, aim4)) { isAccessed[row][aim4] = true; back(time + killTime, row, aim4); isAccessed[row][aim4] = false; } break; } } }
BFS代码
int bfs() { int time = 0; while (true) { time++; //对当前点每个元素遍历，查看是否能生成意愿点 for (int i = 0; i < current.size(); i ++) { Point p = current.get(i); int row1 = (int) p.getX(); int col1 = (int) p.getY(); //debug 这里的横坐标 0010101010101010 why？ if (row1 == height - 1 && col1 == width - 1) { return time + 1; } //如果当前点还有怪物时间，你就不能对任何点产生企图 if (flag[row1][col1] > 0) { flag[row1][col1]--; if (flag[row1][col1] == 0) { flag[row1][col1] = -1;//时间消费光了，标记数组为已走 } //产生企图,生成will集合,标记数组 } else if (flag[row1][col1] == -1) { //生成意愿 aim(row1 + 1, col1); aim(row1 - 1, col1); aim(row1, col1 + 1); aim(row1, col1 + 1); //debug 第二次取就是空的了,很有可能是之后的步骤错了 } } //debug will的size第一次是1，之后一直是0了 1,0 if (will.size() != 0) { for (int j = 0; j < will.size(); j++) { Point p1 = will.get(j); int row2 = (int) p1.getX(); int col2 = (int) p1.getY(); //如果有怪物 if (!a[row2][col2].equals(".")) { int num = Integer.valueOf(a[row2][col2]); flag[row2][col2] = num; } flag[row2][col2] = -1; current.add(p1); } will.clear(); } } }
全部代码
package ACM1000_1099; import java.awt.*; import java.util.ArrayList; import java.util.List; import java.util.Queue; public class IgnatiusAndThePrincessOne1026 { IgnatiusAndThePrincessOne1026(String[][] a) { height = a.length; width = a[0].length; this.a = new String[height][width]; isAccessed = new boolean[height][width]; for (int i = 0; i < height; i++) { for (int j = 0; j < width; j++) { this.a[i][j] = a[i][j]; } } min = Integer.MAX_VALUE; // back(-1, 0, 0); // // System.out.print(min + ""); //0 未走 -1 已走 -2将走 正数 剩余时间//所以当前点有两种情况 -1已走，正数 有倒计时 flag = new int[height][width]; current = new ArrayList<>(); will = new ArrayList<>(); flag[0][0] = -1; Point point = new Point(0, 0); current.add(point); int result = bfs(); System.out.print(result + ""); } String[][] a; boolean[][] isAccessed; int height; int width; int min; int[][] flag; Listcurrent; List will; boolean isAccessable(int row, int col) { if (row < 0 || col < 0 || row > height - 1 || col > width - 1 || a[row][col].equals("X") || isAccessed[row][col]) { return false; } return true; } //回溯 即 dfs？ void back(int time, int row, int col) {//origin time = 0//row 行 // col 列 int killTime = 1; if (!a[row][col].equals(".")) { killTime += Integer.valueOf(a[row][col]); } if (col == width - 1 && row == height - 1) { min = Math.min(min, killTime + time); return; } for (int i = 0; i < 4; i++) { switch (i) { case 0: int aim = row + 1; if (isAccessable(aim, col)) { isAccessed[aim][col] = true; back(time + killTime, aim, col); isAccessed[aim][col] = false; } break; case 1: int aim2 = row - 1; if (isAccessable(aim2, col)) { isAccessed[aim2][col] = true; back(time + killTime, aim2, col); isAccessed[aim2][col] = false; } break; case 2: int aim3 = col + 1; if (isAccessable(row, aim3)) { isAccessed[row][aim3] = true; back(time + killTime, row, aim3); isAccessed[row][aim3] = false; } break; case 3: int aim4 = col - 1; if (isAccessable(row, aim4)) { isAccessed[row][aim4] = true; back(time + killTime, row, aim4); isAccessed[row][aim4] = false; } break; } } } boolean isWillable(int row, int col) { if (row < 0 || col < 0 || row > height - 1 || col > width - 1 || a[row][col].equals("X") || isAccessed[row][col]) { return false; } if (flag[row][col] != -1 && flag[row][col] != -2) { return true; } else { return false; } } void aim(int row, int col) { //如果可产生企图 if (isWillable(row, col)) { flag[row][col] = -2; Point p = new Point(row, col); will.add(p); } } //优先队列还不是太明白 但是bfs可以试一下 int bfs() { int time = 0; while (true) { time++; //对当前点每个元素遍历，查看是否能生成意愿点 for (int i = 0; i < current.size(); i ++) { Point p = current.get(i); int row1 = (int) p.getX(); int col1 = (int) p.getY(); //debug 这里的横坐标 0010101010101010 why？ if (row1 == height - 1 && col1 == width - 1) { return time + 1; } //如果当前点还有怪物时间，你就不能对任何点产生企图 if (flag[row1][col1] > 0) { flag[row1][col1]--; if (flag[row1][col1] == 0) { flag[row1][col1] = -1;//时间消费光了，标记数组为已走 } //产生企图,生成will集合,标记数组 } else if (flag[row1][col1] == -1) { //生成意愿 aim(row1 + 1, col1); aim(row1 - 1, col1); aim(row1, col1 + 1); aim(row1, col1 + 1); //debug 第二次取就是空的了,很有可能是之后的步骤错了 } } //debug will的size第一次是1，之后一直是0了 1,0 if (will.size() != 0) { for (int j = 0; j < will.size(); j++) { Point p1 = will.get(j); int row2 = (int) p1.getX(); int col2 = (int) p1.getY(); //如果有怪物 if (!a[row2][col2].equals(".")) { int num = Integer.valueOf(a[row2][col2]); flag[row2][col2] = num; } flag[row2][col2] = -1; current.add(p1); } will.clear(); } } } public static void main(final String[] args) throws Exception { String[][] a = { {".", "X", "X", ".", "1", "."}, {".", ".", "X", ".", "2", "."}, {"2", ".", ".", ".", "X", "."}, {".", ".", ".", "X", "X", "."}, {"X", "X", "X", "X", "X", "."}}; IgnatiusAndThePrincessOne1026 i = new IgnatiusAndThePrincessOne1026(a); } }
这篇文章写的真烂！但是今天状态不好。虽然这dfs bfs很快就写出来了。