Input
Output
Sample Input
2 1 #. .# 4 4 ...# ..#. .#.. #... -1 -1
Sample Output
2 1
解题思路:
回溯法递归
从第一行开始每一个一个一个试,下一行也是一个一个试。
AC代码。
import java.util.Scanner; /* * poj 1321 */ public class Main{ static char[][] graph; static boolean[] rows; static boolean[] cols; static int n,k,nums = 0,res = 0; public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(true) { n=sc.nextInt(); k=sc.nextInt(); if(n==-1) break; if(n==1) { System.out.println(1); continue; } graph = new char[n][n]; for(int i=0;i<n;i++) { String str = sc.next(); for(int j=0;j<n;j++) { graph[i][j]=str.charAt(j); } } cols=new boolean[n]; next(0); System.out.println(res); res=0; } } public static void next(int row) { if(row==n) return; for(int i=0;i<n;i++) if(graph[row][i]==‘#‘&&cols[i]==false) { cols[i]=true; nums++; if(k==nums) { res++; } next(row+1); cols[i]=false; nums--; } next(row+1); } }
Description
Input
Output
Escaped in x minute(s).
Trapped!
Sample Input
3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 0 0 0
Sample Output
Escaped in 11 minute(s). Trapped!
解题思路:BFS 注意千万在重新调用之前把数据清干净。
import java.util.HashSet; import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; public class Main{ static int L,R,C,res=-1; static char[][][] graph; static class Node{ int l,r,c; } static Queue<Integer> que = new LinkedList<Integer>(); static HashSet<Integer> hs = new HashSet<Integer>(); public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(true) { L=sc.nextInt(); R=sc.nextInt(); C=sc.nextInt(); if(L==0)break; graph = new char[L][R][C]; for(int i=0;i<L;i++)for(int j=0;j<R;j++) { String str = sc.next(); for(int k=0;k<C;k++) { graph[i][j][k]=str.charAt(k); if(graph[i][j][k]==‘S‘) { hs.add(set(i,j,k)); que.offer(set(i,j,k)); } } } bfs(); res=-1; hs.clear(); que.clear(); } } public static void bfs() { res++; int size = que.size(); if(size==0) { System.out.println("Trapped!"); return; } while(size-->0){ Node node = get(que.poll()); if(graph[node.l][node.r][node.c]==‘E‘) { System.out.println("Escaped in "+res+" minute(s)."); return; } //上 if(node.l!=L-1) { if(graph[node.l+1][node.r][node.c]==‘.‘&&!hs.contains(set(node.l+1,node.r,node.c))) { hs.add(set(node.l+1,node.r,node.c)); que.offer(set(node.l+1,node.r,node.c)); }else if(graph[node.l+1][node.r][node.c]==‘E‘) { que.offer(set(node.l+1,node.r,node.c)); } } //下 if(node.l!=0) { if(graph[node.l-1][node.r][node.c]==‘.‘&&!hs.contains(set(node.l-1,node.r,node.c))) { hs.add(set(node.l-1,node.r,node.c)); que.offer(set(node.l-1,node.r,node.c)); }else if(graph[node.l-1][node.r][node.c]==‘E‘) { que.offer(set(node.l-1,node.r,node.c)); } } //左 if(node.r!=0) { if(graph[node.l][node.r-1][node.c]==‘.‘&&!hs.contains(set(node.l,node.r-1,node.c))) { hs.add(set(node.l,node.r-1,node.c)); que.offer(set(node.l,node.r-1,node.c)); }else if(graph[node.l][node.r-1][node.c]==‘E‘) { que.offer(set(node.l,node.r-1,node.c)); } } //右 if(node.r!=R-1) { if(graph[node.l][node.r+1][node.c]==‘.‘&&!hs.contains(set(node.l,node.r+1,node.c))) { hs.add(set(node.l,node.r+1,node.c)); que.offer(set(node.l,node.r+1,node.c)); }else if(graph[node.l][node.r+1][node.c]==‘E‘) { que.offer(set(node.l,node.r+1,node.c)); } } //前 if(node.c!=0) { if(graph[node.l][node.r][node.c-1]==‘.‘&&!hs.contains(set(node.l,node.r,node.c-1))) { hs.add(set(node.l,node.r,node.c-1)); que.offer(set(node.l,node.r,node.c-1)); }else if(graph[node.l][node.r][node.c-1]==‘E‘) { que.offer(set(node.l,node.r,node.c-1)); } } //后 if(node.c!=C-1) { if(graph[node.l][node.r][node.c+1]==‘.‘&&!hs.contains(set(node.l,node.r,node.c+1))) { hs.add(set(node.l,node.r,node.c+1)); que.offer(set(node.l,node.r,node.c+1)); }else if(graph[node.l][node.r][node.c+1]==‘E‘) { que.offer(set(node.l,node.r,node.c+1)); } } } bfs(); } public static int set(int l,int r,int c) { return l*10000+r*100+c; } public static Node get(int i) { Node node = new Node(); node.l=i/10000; node.r=(i-node.l*10000)/100; node.c=(i-node.l*10000-node.r*100); return node; } }
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Output
Sample Input
5 17
Sample Output
4
Hint
import java.util.LinkedList; import java.util.Queue; import java.util.Scanner; public class Main { static int N,K; static int res=-1; static Queue<Integer> que = new LinkedList<Integer>(); static boolean[] vis = new boolean[1000000]; public static void main(String[] args) { Scanner sc = new Scanner(System.in); N=sc.nextInt(); K=sc.nextInt(); que.offer(N); bfs(0); System.out.println(res); } static void bfs(int deep) { int size = que.size(); if(que.contains(K)) { res=deep; return; } while(size-->0) { int P = que.poll(); vis[P]=true; if(P<K) { if(lim(P+1)&&!vis[P+1])que.offer(P+1); if(lim(P*2)&&!vis[P*2])que.offer(P*2); if(lim(P-1)&&!vis[P-1])que.offer(P-1); }else { if(lim(P-1)&&!vis[P-1])que.offer(P-1); } } if(que.size()!=0)bfs(deep+1); } static boolean lim(int A) { if(A>100000||A<0) return false; else return true; } }
Description
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored black on one side and white on the other side.
As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.
Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word "IMPOSSIBLE".
Input
Output
Sample Input
4 4 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1
Sample Output
0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0
思路:二进制枚举+自上而下遍历。
WA错误:要求反转次数最小、其次是字典序最小。(MD挂了好些次)
import java.util.Arrays; import java.util.HashMap; import java.util.Scanner; public class Main{ static int[][] graph; static int N, M; public static void main(String[] args) { Scanner sc = new Scanner(System.in); M = sc.nextInt(); N = sc.nextInt(); if (N == 0) return; graph = new int[M][N]; int[][] meijushu = setFirst(); for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { graph[i][j] = sc.nextInt(); } } int MinRes = Integer.MAX_VALUE; HashMap<Integer, int[][]> hm = new HashMap<Integer,int[][]>(); for (int mjs = 0; mjs < meijushu.length; mjs++) {// 枚举数组 int[][] newGraph = new int[M][N]; int[][] res = new int[M][N]; for (int i = 0; i < M; i++) { newGraph[i] = Arrays.copyOf(graph[i], N); } // 第一行下棋 res[0] = Arrays.copyOf(meijushu[mjs], N); for (int i = 0; i < N; i++) { if (meijushu[mjs][i] == 1) chess(newGraph, 0, i); } // 接下来每行下棋 for (int i = 1; i < M; i++) { for (int j = 0; j < N; j++) { if (newGraph[i - 1][j] == 1) { chess(newGraph, i, j); res[i][j] = 1; } } } int sign = 0; for (int i = 0; i < N; i++) { if (newGraph[M - 1][i] == 1) { sign++; break; } } if (sign == 0) { int a = 0; for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { if (1 == res[i][j]) a++; } } if (a < MinRes) { MinRes = a; hm.put(a, res); } } } if (!hm.isEmpty()) { int[][] res = hm.get(MinRes); for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) { System.out.print(res[i][j] + " "); } System.out.println(); } return; } System.out.println("IMPOSSIBLE"); } // 二进制枚举数组 public static int[][] setFirst() { int[][] newGraph = new int[(int) Math.pow(2, N)][N]; String str[] = new String[(int) Math.pow(2, N)]; for (int i = 0; i < (int) Math.pow(2, N); i++) { str[i] = Integer.toBinaryString(i); int len = str[i].length(); for (int j = 0; j < N - len; j++) { str[i] = "0" + str[i]; } } for (int i = 0; i < (int) Math.pow(2, N); i++) { for (int j = N - 1; j >= 0; j--) { newGraph[i][j] = str[i].charAt(j) - 48; } /* * for(int j=0;j<N;j++) { System.out.print(newGraph[i][j]+"--"); } * System.out.println(); */ } return newGraph; } // 下棋 public static void chess(int[][] newGraph, int x, int y) { if (newGraph[x][y] == 1) { newGraph[x][y] = 0; } else if (newGraph[x][y] == 0) { newGraph[x][y] = 1; } // 左 if (x > 0) { if (newGraph[x - 1][y] == 1) { newGraph[x - 1][y] = 0; } else if (newGraph[x - 1][y] == 0) { newGraph[x - 1][y] = 1; } } // 上 if (y > 0) { if (newGraph[x][y - 1] == 1) { newGraph[x][y - 1] = 0; } else if (newGraph[x][y - 1] == 0) { newGraph[x][y - 1] = 1; } } // 下 if (y < N - 1) { if (newGraph[x][y + 1] == 1) { newGraph[x][y + 1] = 0; } else if (newGraph[x][y + 1] == 0) { newGraph[x][y + 1] = 1; } } // 右 if (x < M - 1) { if (newGraph[x + 1][y] == 1) { newGraph[x + 1][y] = 0; } else if (newGraph[x + 1][y] == 0) { newGraph[x + 1][y] = 1; } } } }
原文:https://www.cnblogs.com/godoforange/p/11202906.html