P1123 取数游戏 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
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本题一开始是想写双for循环，复杂度\(O(n^n)\)，不出意外T了三个点，然后思考剪枝，考虑到不需要重新找，做了一个bool数组来标记，防止重新搜索已经出现过的方案，过了第四个点。
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后面看了题解后，发现这一思路极好，枚举整个棋盘，每个点可以选也可以不选，则复杂度直接降为\(O(2^n)\),直接ac，根本不用剪枝
package shuati;import java.io.*;
import java.util.*;public class Main {static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static Scanner scanner = new Scanner(System.in);static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));static int t, n, m;static final int N = 10;static int[][] g = new int[N][N];// cnt[i][j] 表示以（i，j）为中心的周围八个格子有几个数被选取static int[][] cnt = new int[N][N];static int ans, nowans;static int dx[] = {0, 1, 1, 0, -1, -1, -1, 0, 1};static int dy[] = {0, 0, 1, 1, 1, 0, -1, -1, -1};public static void dfs(int x, int y) {// 处理越界if(y == m + 1) {dfs(x + 1, 1);return;}if(x == n + 1) {ans = Math.max(nowans, ans);return;}// 不选择这个点dfs(x,y + 1);if(cnt[x][y] == 0) {for(int i = 0; i < 9; i++) {cnt[x + dx[i]][y + dy[i]]++;}nowans += g[x][y];dfs(x, y + 1);nowans -= g[x][y];for(int i = 0; i < 9; i++) {cnt[x + dx[i]][y + dy[i]]--;}}}public static void solve() throws IOException {for (int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {cin.nextToken();g[i][j] = (int) cin.nval;}}dfs(1, 1);cout.println(ans);cout.flush();}public static void main(String[] args) throws IOException{cin.nextToken();t = (int) cin.nval;while(t-- != 0) {cin.nextToken();n = (int) cin.nval;cin.nextToken();m = (int) cin.nval;ans = 0;nowans = 0;solve();}}
}
深さ優先探索 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
- 经典搜索
import java.io.*;
import java.util.*;public class Main {static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static Scanner scanner = new Scanner(System.in);static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));static final int N = 510;static int n, m;//长和宽static char[][] g = new char[N][N];static boolean[][] str = new boolean[N][N];static int[] dx = {1, 0, -1, 0};static int[] dy = {0, 1, 0, -1};static int cnt;public static void dfs(int x, int y) {if(g[x][y] == 'g') {cnt = 1;cout.print("Yes");return;}for(int i = 0; i < 4; i++) {int nowx = dx[i] + x;int nowy = dy[i] + y;if(nowx >= 1 && nowx <= n && nowy >= 1 && nowy <= m) {if(str[nowx][nowy] == false && g[nowx][nowy] != '#' && cnt == 0) {str[nowx][nowy] = true;dfs(nowx, nowy);}}}}public static void main(String[] args) throws IOException{
// cin.nextToken();
// n = (int) cin.nval;
// cin.nextToken();
// m = (int) cin.nval;n = scanner.nextInt();m = scanner.nextInt();int chux = 0;int chuy = 0;for(int i = 1; i <= n; i++) {String s = scanner.next();for(int j = 1; j <= m; j++) {g[i][j] = s.charAt(j - 1);if(s.charAt(j - 1) == 's') {chux = i;chuy = j;}}}str[chux][chuy] = true;dfs(chux, chuy);if(cnt == 0) {cout.print("No");}cout.flush();}
}
P1706 全排列问题 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)
- 水题
import java.io.*;
import java.util.*;public class Main {static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));static int n;static boolean[] str = new boolean[10];static Queue<Integer> myQueue = new LinkedList<>();public static void dfs(int cnt) {if(cnt == n) {for (Integer i : myQueue) {cout.print(" " + i);}cout.println();cout.flush();return;}for(int i = 1; i <= n; i++) {if(str[i] == false) {str[i] = true;cnt++;myQueue.add(i);dfs(cnt);myQueue.remove(i);cnt--;str[i] = false;}}}public static void main(String[] args) throws IOException{cin.nextToken();n = (int)cin.nval;dfs(0);}
}
