1.17 刷题

1 思路

P1331 海战 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 本题难点主要是如何分辨哪些穿是相撞而产生无效，哪些是有效
• 很容易想到的是，不论是bfs还是dfs都可以轻松全部搜掉，只需要简单的遍历所有点，然后套板子即可
• 但是这是无法排除无效情况的，也就是相撞的情况
• 推敲一下，发现其实就只有4种情况会无效

• 那么只要写个判断函数就可以排除这四种情况
• 即只需要dfs搜索+判断即可解决这道题

2 代码

#include<iostream>
#include<map>
#include<cstdio>
#include<cstring>
using namespace std;const int N = 1e3 + 10;
int m[N][N];
bool str[N][N];
int r, c, ans;
int dx[] = {0, 1, 1};
int dy[] = {1, 0, 1};int Judge(int x, int y) {int d1 = m[x][y];int d2 = m[x + 1][y];int d3 = m[x][y + 1];int d4 = m[x + 1][y + 1];// for(int i = 0; i < 2; i++) {//    for(int j = 0; j < 2; j++) {//       int nowx = x + dx[i];//       int nowy = y + dy[i];//       if(nowx <= c && nowy <= r) {//          if(m[nowx][nowy] == 1) {//             str[nowx][nowy] = 1;//          } else {//             return -1;//          }//       }//    }// }if(d1 == 1 && d2 == 1 && d3 == 1 && d4 == 0) return -1;if(d1 == 1 && d4 == 1 && d3 == 1 && d2 == 0) return -1;if(d1 == 1 && d2 == 1 && d4 == 1 && d3 == 0) return -1;if(d4 == 1 && d2 == 1 && d3 == 1 && d1 == 0) return -1;return 1;
}int po() {for(int i = 1; i <= r; i++) {for(int j = 1; j <= c; j++) {int pan = Judge(i,j);if(pan == -1) {cout<<"Bad placement."<<endl;return -1;}}}return 0;
}void dfs(int x, int y) {for(int i = 0; i < 3; i++) {int nowx = x + dx[i];int nowy = y + dy[i];if(m[nowx][nowy] == 1 && str[nowx][nowy] == false) {str[nowx][nowy] = true;dfs(nowx, nowy);}}
}int main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);memset(m, 0, sizeof(m));memset(str, false, sizeof(str));cin>>r>>c;for(int i = 1;i <= r; i++) {for(int j = 1;j <= c; j++) {char s;cin>>s;if(s == '#') {m[i][j] = 1;}}}int k = po();if(k == -1) return 0;memset(str, false, sizeof(str));for(int i = 1;i <= r; i++) {for(int j = 1;j <= c; j++) {if(m[i][j] == 1 && str[i][j] == false) {str[i][j] = true;dfs(i, j);ans++;}}}cout<<"There are "<<ans<<" ships.";return 0;}

import java.util.Scanner;public class Main {static final int N = 1010;static int[][] m = new int[N][N];static boolean[][] str = new boolean[N][N];static int r, c, ans;static int[] dx = {0, 1, 1};static int[] dy = {1, 0, 1};public static int Judge(int x, int y) {int d1 = m[x][y];int d2 = m[x + 1][y];int d3 = m[x][y + 1];int d4 = m[x + 1][y + 1];if(d1 == 1 && d2 == 1 && d3 == 1 && d4 == 0) return -1;if(d1 == 1 && d4 == 1 && d3 == 1 && d2 == 0) return -1;if(d1 == 1 && d2 == 1 && d4 == 1 && d3 == 0) return -1;if(d4 == 1 && d2 == 1 && d3 == 1 && d1 == 0) return -1;return 1;}public static int po() {for(int i = 1; i <= r; i++) {for(int j = 1; j <= c; j++) {int pan = Judge(i,j);if(pan == -1) {System.out.println("Bad placement.");return -1;}}}return 0;}public static void dfs(int x, int y) {for(int i = 0; i < 3; i++) {int nowx = x + dx[i];int nowy = y + dy[i];if(m[nowx][nowy] == 1 && str[nowx][nowy] == false) {str[nowx][nowy] = true;dfs(nowx, nowy);}}}public static void main(String[] args) {Scanner scanner = new Scanner(System.in);r = scanner.nextInt();c = scanner.nextInt();for(int i = 1;i <= r; i++) {String s1 = scanner.next();for(int j = 1;j <= c; j++) {char s;s = s1.charAt(j - 1);if(s == '#') {m[i][j] = 1;}}}int k = po();if(k == -1) return;for(int i = 1;i <= r; i++) {for(int j = 1;j <= c; j++) {if(m[i][j] == 1 && str[i][j] == false) {str[i][j] = true;dfs(i, j);ans++;}}}System.out.println("There are " + ans + " ships.");return;}}