蓝桥杯-最大联通问题以及全球变暖

110010000011111110101001001001101010111011011011101001111110

010000000001010001101100000010010110001111100010101100011110 

001011101000100011111111111010000010010101010111001000010100 

101100001101011101101011011001000110111111010000000110110000 

010101100100010000111000100111100110001110111101010011001011 

010011011010011110111101111001001001010111110001101000100011 

101001011000110100001101011000000110110110100100110111101011 

101111000000101000111001100010110000100110001001000101011001 

001110111010001011110000001111100001010101001110011010101110 

001010101000110001011111001010111111100110000011011111101010 

011111100011001110100101001011110011000101011000100111001011 

011010001101011110011011111010111110010100101000110111010110 

001110000111100100101110001011101010001100010111110111011011 

111100001000001100010110101100111001001111100100110000001101 

001110010000000111011110000011000010101000111000000110101101 

100100011101011111001101001010011111110010111101000010000111 

110010100110101100001101111101010011000110101100000110001010 

110101101100001110000100010001001010100010110100100001000011 

100100000100001101010101001101000101101000000101111110001010 

101101011010101000111110110000110100000010011111111100110010 

101111000100000100011000010001011111001010010001010110001010 

001010001110101010000100010011101001010101101101010111100101 

001111110000101100010111111100000100101010000001011101100001 

101011110010000010010110000100001010011111100011011000110010 

011110010100011101100101111101000001011100001011010001110011 

000101000101000010010010110111000010101111001101100110011100 

100011100110011111000110011001111100001110110111001001000111 

111011000110001000110111011001011110010010010110101000011111 

011110011110110110011011001011010000100100101010110000010011 

010011110011100101010101111010001001001111101111101110011101

 代码（DFS方法)

#include <iostream>
#include <cmath>
using namespace std;
int dx[]={-1,1,0,0},dy[]={0,0,-1,1};  //方向数组存储四个方向坐标偏移量 
char g[35][65];
int res=1;
int dfs(int x,int y){
    int cnt=1;
	g[x][y]='0';      //每次将该位置的1标记为已搜过 
	for(int i=0;i<4;i++){
		int a=x+dx[i],b=y+dy[i];//顺序为左，右，下，上 
		if(a>=0&&a<30&&b>=0&&b<60&&g[a][b]=='1'){    
			cnt+=dfs(a,b);             //深搜统计1的个数 
		} 
	}
	return cnt;
}
int main(){
    for(int i=0;i<30;i++){
    	for(int j=0;j<60;j++){
			cin>>g[i][j];
		}
	}
    for(int i=0;i<30;i++){
    	for(int j=0;j<60;j++){
    		if(g[i][j]=='1'){       //枚举每个1的位置，进行深搜，答案即为某个1深搜得到的1的总数的最大值 
    			res=max(dfs(i,j),res);
			}
		}
	}
    cout<<res;
	return 0;
}

全球变暖

#include<bits/stdc++.h>
using namespace std;

int n;
char a[1010][1010]; //地图
int vis[1010][1010]={0};  //标记是否搜过
int d[4][2] = {{0,1}, {0,-1}, {1,0}, {-1,0}}; //四个方向，也可以这样去标记 
int flag;  //用于标记这个岛中是否被完全淹没
void dfs(int x, int y){
    vis[x][y] = 1;      //标记这个'#'被搜过。注意为什么可以放在这里
    if(a[x][y+1]=='#' && a[x][y-1]=='#' && a[x+1][y]=='#' && a[x-1][y]=='#')
        flag = 1;       //上下左右都是陆地，不会淹没
    for(int i = 0; i < 4; i++){ //继续DFS周围的陆地
        int nx = x + d[i][0], ny = y + d[i][1];
      //if(nx>=1 && nx<=n && ny>=1 && ny<=n && vis[nx][ny]==0 && a[nx][ny]=='#') //题目说边上都是水，所以不用这么写了
        if(vis[nx][ny]==0 && a[nx][ny]=='#') //继续DFS未搜过的陆地，目的是标记它们
            dfs(nx,ny);
    }
}

int main(){
    cin >> n;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            cin >> a[i][j];
    int ans = 0 ;
    for(int i = 1; i <= n; i++)  //DFS所有像素点
        for(int j = 1; j <= n; j++)
            if(a[i][j]=='#' && vis[i][j]==0){
                flag = 0;
                dfs(i,j);
                if(flag == 0)  //这个岛全部被淹
                    ans++;     //统计岛的数量
            }
    cout<<ans<<endl;
    return 0;
}