亚特兰蒂斯

答案为 661 (搜索是蓝桥常考算法)

从任意一个角出发搜索整张图 , 我们需要维护两个变量 cnt 代表从 (0,0) 出发是否能到达其他三个角(0,29),(29,0),(29,29) ;还有ans表示答案个数

如果存在被彻底淹没的土地，那么四个角一定是相连通的(任意两点之间都存在路径可以相互到达)

假设(x,y)点会被彻底淹没那么从(x,y)这点原路返回一定可以回归四个角,那么从任意一个角出发一定可以找到路径 去往其他三个角 ( 通过(x,y)作为中间点 )。

那么我们就可以逮着其中一个点(0,0)点开始搜索并记录一共可以到达哪些为 '0' 的点

BFS做法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 35 ;

    char g[50][50] = {
"001000000001000010001001000000",
"001000001101011010010000000111",
"100000110000100101100000100000",
"000001000100100000100000000001",
"010001000010001100100010000010",
"000000101101000010100100010001",
"010000100000010000001011000000",
"010001010011101100100001100100",
"100000100110000000000010001001",
"000000010011100011010101000000",
"100101100011000000001001101000",
"100000010010000100000100000000",
"010000100000001100000011000001",
"010000100010000001000001001100",
"000001101010110001000000100000",
"000000000000101000000100100001",
"001010100111010100000010000010",
"000000100100000000000100000000",
"110011001000000000010000110011",
"000001010010000010011100001000",
"001011001011101010001000000001",
"011000000100000100000100101100",
"000100011100010111000100001011",
"000010000000011001000000001010",
"001000100100000000000100000000",
"110000000101010101110110011110",
"100001010001000000101100100011",
"011001000100001100000000000000",
"000000000000000000000000000000",
"000000000000000000000000000000"} ; 

int n , m , cnt , ans  ; 
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

bool st[N][N] ; 

struct node
{
    int x , y ;         
}q[N * N ] ; 

void bfs()
{
    int hh = 0 , tt = - 1 ; 
    q[ ++ tt ] = {0 , 0 } ; 
    st[0][0] = true ; 
    
    while(hh <= tt ) 
    {
        auto t = q[hh ++ ] ; //  判断是否能到达其余三个角
        if( (t.x == 0 and t.y == m - 1 ) or (t.x == n - 1 and t.y == 0 ) or (t.x == n - 1 and t.y == m - 1 ) ) cnt ++ ; 
        ans ++ ; 
		for(int i = 0 ; i < 4 ; i ++ ) 
        {
            int x = dx[i] + t.x , y = dy[i] + t.y ; 
            if(x < 0 or x >= n or y < 0 or y >= m or g[x][y] == '1' )continue ; 
            if(!st[x][y]) 
            {
                st[x][y] = true ; 
                q[++ tt ] = {x , y} ; 
            }
        }
    }
}

int main()
{
    n = 30 , m = 30 ; 
    
    bfs() ; 
    
    int c = 0 ; 
    
    //如果 从 (0,0) 出发可以到达 其他三个 点说明 这张地图一定是能够汇聚四方洪水的 
    if(cnt == 3)cout << ans << endl ; // 此时答案就是从任意一个角搜索 所能蔓延到的 所有格子数量 
    else cout << 0 << endl ;   
    
    return 0 ; 
}