bfs36分求助

P8662 [蓝桥杯 2018 省 AB] 全球变暖

zhz_aleph_114514 @ 2022-11-30 19:04:25

code:

#include<bits/stdc++.h>
#include<queue>
#define ll long long
#define ull unsigned long long
using namespace std;
struct node{
    int x,y;
};
char mapp[1005][1005];
char tmp[1005][1005];
int n;
queue<node> q;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
void bfs(int x6,int y6){
    q.push(node{x6,y6});
    mapp[x6][y6]='$';
    int kx,ky;
    int xx,yy;
    while(!q.empty()){
        node kkk=q.front();
        q.pop();
        kx=kkk.x; ky=kkk.y;
        for(int i=0;i<4;i++){
            xx=kx+dx[i];
            yy=ky+dy[i];
            if(mapp[xx][yy]=='#'&&xx>=0&&yy>=0&&xx<n&&yy<n){
                mapp[xx][yy]='$';
                q.push(node{xx,yy});
            }
        }
    }
}
void bfs1(int x6,int y6){
    q.push(node{x6,y6});
    tmp[x6][y6]='$';
    int kx,ky;
    int xx,yy;
    while(!q.empty()){
        node kkk=q.front();
        q.pop();
        kx=kkk.x; ky=kkk.y;
        for(int i=0;i<4;i++){
            xx=kx+dx[i];
            yy=ky+dy[i];
            if(tmp[xx][yy]=='#'&&xx>=0&&yy>=0&&xx<n&&yy<n){
                tmp[xx][yy]='$';
                q.push(node{xx,yy});
            }
        }
    }
}
int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            cin>>mapp[i][j];
            tmp[i][j]=mapp[i][j];
        }
    }
    int t0=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(tmp[i][j]=='#'){
                t0++;
                bfs1(i,j);
            }
        }
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(mapp[i][j]=='#'){
                if(mapp[i+1][j]=='.'||mapp[i][j+1]=='.'||mapp[i-1][j]=='.'||mapp[i][j-1]=='.'){
                    mapp[i][j]='$';
                }
            }
        }
    }
    int ans=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(mapp[i][j]=='#'){
                ans++;
                bfs(i,j);
            }
        }
    }
    if(ans>=t0){
        cout<<0;
        return 0;
    }  
    cout<<t0-ans;
    return 0;
}

提交记录


by zhz_aleph_114514 @ 2022-11-30 19:06:11

思路:开始先bfs求出岛屿数量;之后双重for循环模拟淹没岛屿;最后bfs求出未被淹没的岛屿数量,最后相减得出答案

(不确定自己的思路对不对


by jnyz2021109122116 @ 2022-11-30 22:32:41

@zhz_aleph_114514 个人拙见这样判断岛屿数量可能会有bug
如果一个岛屿被完全淹没而另一个岛屿被分割成了两个甚至更多,那么此时被淹没岛屿数量显然至少为1.


by jnyz2021109122116 @ 2022-11-30 22:36:24

比如说这样

7 
.......
.##.##.
.#####.
.##.##.
......
...##..
.......  

|