1.ACW1097: Pond count
Title:'W': Water,'.': Dry, find the number of puddles.
Idea: The most basic Floodfill.
The code is as follows:

#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
const int N = 1010;
typedef long long LL;
typedef pair<LL, LL> PII;
char c[N][N];
LL n,m;
bool st[N][N];
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void bfs(LL sta,LL ed)
{
    queue<PII>q;q.push({sta,ed});st[sta][ed]=true;
    while(q.size())
    {
        PII t=q.front();q.pop();
        for(int i=0;i<8;i++)
        {
            LL ix=t.x+dx[i],iy=t.y+dy[i];
            if(ix<=0||iy<=0||ix>n||iy>m||st[ix][iy]||c[ix][iy]=='.') continue;
            q.push({ix,iy});st[ix][iy]=true;
        }
    }
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++) cin>>c[i][j];
    LL ans=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(!st[i][j]&&c[i][j]=='W')
            {
                bfs(i,j);
                ans++;
            }
    cout<<ans;
    return 0;
}

2.ACW1098: The Castle Problem
Inscription: Each grid has a number indicating which direction of the four directions there is a wall. Two cells without a wall are connected to a room to find the maximum number of rooms. Among them, 1: West Wall, 2: North Wall, 3: East Wall, 4: South Wall, [can be added up, for example, if there are four walls, this grid is 15.)
Idea: It is unnecessary to build a graph and find out if every bit of a grid is 1 by binary judgment. A 1 means that the grid has a wall in its current corresponding orientation, so it is not connected to the next-wall grid structure in this direction and will not be added to the queue naturally.
The code is as follows:

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#define x first
#define y second
using namespace std;
const int N = 55;
typedef long long LL;
typedef pair<int, int> PII;
LL c[N][N],n,m,t,res;
bool st[N][N];
int dx[4]={0,-1,0,1},dy[4]={-1,0,1,0};
void bfs(LL sta,LL ed)
{
    queue<PII>q;q.push({sta,ed});st[sta][ed]=true;LL area=0;
    while(q.size())
    {
        PII t=q.front();q.pop();area++;
        for(int i=0;i<4;i++)
        {
            LL ix=t.x+dx[i],iy=t.y+dy[i];
            if(ix<1||iy<1||ix>n||iy>m||st[ix][iy]) continue;
            if(!(c[t.x][t.y]>>i&1)) {q.push({ix,iy});st[ix][iy]=true;}
        }
    }
    res=max(area,res);
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++) cin>>c[i][j];
    LL ans=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(!st[i][j]) {bfs(i,j);ans++;}
    cout<<ans<<endl;
    cout<<res;
    return 0;
}

3.ACW1106: Peaks and Valleys
Topic:
We define a grid whose set S is a mountain (valley) if and only if:
(1) All the lattices of S have the same height.
(2) All the lattices of S are connected.
(3) For s belongs to S, the s'adjacent to s does not belong to S, and there are ws>ws' (mountain peak) or ws<ws'(valley).
(4) If there is no adjacent area around it, consider it both a mountain and a valley.
Idea: Floodfill finds each connected block and marks the points it is adjacent to as large and as small as possible:
If there are any larger or smaller lattices, the connection block is neither a mountain nor a valley.
(2) If there is only one larger than it, it cannot be smaller: the connecting block is a valley.
(3) If only smaller than it is, it cannot be larger: the connecting block is a mountain peak.
(4) If it is neither so large nor so small, it is equal to all the lattices in the current connected block: those lattices that are equal to the connected block are also part of the connected block.
The code is as follows:

#include <bits/stdc++.h>
#define x first
#define y second
using namespace std;
const int N = 1010;
typedef long long LL;
typedef pair<int, int> PII;
LL c[N][N],n,m,t,low,peak;
bool st[N][N];
int dx[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
void bfs(LL x,LL y,bool &has_high,bool &has_low)
{
    queue<PII>q;q.push({x,y});st[x][y]=true;
    while(q.size())
    {
        PII t=q.front();q.pop();
        for(int i=0;i<8;i++)
        {
            LL ix=t.x+dx[i],iy=t.y+dy[i];
            if(ix<1||iy<1||ix>n||iy>n) continue;
            if(c[ix][iy]>c[t.x][t.y]) has_low=true;
            else if(c[ix][iy]<c[t.x][t.y]) has_high=true;
            else if(c[ix][iy]==c[t.x][t.y])
            {
                if(!st[ix][iy]) {q.push({ix,iy});st[ix][iy]=true;}
            }
        }
    }
}
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++) cin>>c[i][j];
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            if(!st[i][j])
            {
                bool has_high=false,has_low=false;
                bfs(i,j,has_high,has_low);
                if(!has_low) peak++;
                if(!has_high) low++;
            }
    cout<<peak<<' '<<low;
    return 0;
}