[Usaco nov07] the largest lake

155. [usaco nov07] largest lake

★Input file:lake.inOutput file:lake.outSimple comparison
Time Limit: 1 s memory limit: 128 MB

Translated by cmykrgb123

Description

In a recent rainstorm, Farmer John's farm was drowned, forming lakes. What's more terrible is that his cows are most afraid of water. Fortunately, his insurance company will give him a lot of compensation, which depends on the size of the largest lake on his farm.

His farm can be depicted as a grid of N (1 ≤ n ≤ 100) rows M (1 ≤ m ≤ 100) columns. Each grid is either land or lake water. It is known that there are K (1 ≤ k ≤ n × m) grids with water. Each grid of water can belong to a lake only when its sides are adjacent, and its vertices are not adjacent.

Calculate the size of the largest lake on his farm.

Input

• Row 1st: Three integers: n, m, K
• 2nd. k + 1 row: two integers in the I + 1 row, indicating the location of water in the r row, column C: R, c

Output

• Row 1st: the largest lake size (the number of grids that comprise it)

Sample Input

3 4 53 22 23 12 31 1

Sample output

4

Simple BFs.

#include<cstdio>#include<queue>using namespace std;int map[110][110];bool vis[110][110];int dx[]={0,0,1,-1};int dy[]={1,-1,0,0};int n,m;struct node{    int x,y;}s_pos;int ans;void bfs(int sx,int sy){    queue<node> q;    s_pos.x=sx; s_pos.y=sy;    int sum=1;    vis[sx][sy]=true;    q.push(s_pos);    while(!q.empty()){        node now = q.front();q.pop();        if(sum>ans) ans=sum;        for(int i=0;i<4;i++){            node next = now;            next.x+=dx[i];  next.y+=dy[i];            if(!vis[next.x][next.y]&&map[next.x][next.y]==1){                sum++;                vis[next.x][next.y]=true;                q.push(next);            }        }    }}int main(){    freopen("lake.in","r",stdin);    freopen("lake.out","w",stdout);    int n,m,k;    int a,b;    scanf("%d%d%d",&n,&m,&k);    for(int i=0;i<k;i++){        scanf("%d%d",&a,&b);        map[a][b]=1;    }    for(int i=1;i<=n;i++){       for(int j=1;j<=m;j++)        if(map[i][j]==1&&!vis[i][j]){             bfs(i,j);        }    }    printf("%d\n",ans);    return 0;}