kuangbin专题简单搜索题目几道题目
1、POJ1321棋盘问题
Input
每组数据的第一行是两个正整数，n k，用一个空格隔开，表示了将在一个n*n的矩阵内描述棋盘，以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状：每行有n个字符，其中 # 表示棋盘区域， . 表示空白区域（数据保证不出现多余的空白行或者空白列）。
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 = ,res = ;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(true) {
n=sc.nextInt();
k=sc.nextInt();
if(n==-) break;
if(n==) {
System.out.println();
continue;
}
graph = new char[n][n];
for(int i=;i<n;i++) {
String str = sc.next();
for(int j=;j<n;j++) {
graph[i][j]=str.charAt(j);
}
}
cols=new boolean[n];
next();
System.out.println(res);
res=;
} }
public static void next(int row) {
if(row==n) return;
for(int i=;i<n;i++)
if(graph[row][i]=='#'&&cols[i]==false) {
cols[i]=true;
nums++;
if(k==nums) {
res++;
}
next(row+);
cols[i]=false;
nums--;
}
next(row+);
}
}
2、POJ2251 Dungeon Master
Description
Is an escape possible? If yes, how long will it take?
Input
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.
Output
Escaped in x minute(s).
where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
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;
}
}
3、Catch That Cow
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;
}
}
3、POJ3279Fliptile
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
Lines 2..M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white
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;
}
}
} }
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