F. Avoid The Lakes

                                             F. Avoid The LakesTime Limit:
1000msCase Time Limit:
1000msMemory Limit:
65536KB64-bit integer IO format:
%lld      Java class name:
Main Submit

Status
PID: 3739 Font Size:

Farmer John's farm was flooded in the most recent storm, a fact only aggravated by the information that his cows are deathly afraid of water. His insurance agency will only repay him, however, an amount depending on the size
of the largest "lake" on his farm.The farm is represented as a rectangular grid with
N (1 ≤ N ≤ 100) rows and M (1 ≤ M ≤ 100) columns. Each cell in the grid is either dry or submerged, and exactly
K (1 ≤ K ≤ N × M) of the cells are submerged. As one would expect, a lake has a central cell to which other cells connect by sharing a long edge (not a corner). Any cell that shares a long edge with the central cell or shares
a long edge with any connected cell becomes a connected cell and is part of the lake.

Input

* Line 1: Three space-separated integers: N,
M, and K
* Lines 2..K+1: Line i+1 describes one submerged location with two space separated integers that are its row and column:
R and C

Output

* Line 1: The number of cells that the largest lake contains.　

Sample Input

3 4 53 22 23 12 31 1

Sample Output

4

#include<stdio.h>#include<string.h>#include<queue>using namespace std;int map[102][102];int dir[4][2]= {1,0,0,1,-1,0,0,-1};struct node{    int x;    int y;};int bfs(int x,int y){    int sum=0;    node first,next;    queue<node> q;    first.x=x;    first.y=y;    q.push(first);    map[x][y]=0;    sum=1;    while(!q.empty())    {        first=q.front();        q.pop();        for(int i=0; i<4; i++)        {            next.x=first.x+dir[i][0];            next.y=first.y+dir[i][1];            if(map[next.x][next.y])            {                map[next.x][next.y]=0;                sum++;                q.push(next);            }        }    }    return sum;}int main(){    int n,m,t;    int max1,sum;    int x,y;    int i,j;    while(scanf("%d%d%d",&n,&m,&t)!=EOF)    {        max1=0;        node ste[10000];        memset(map,0,sizeof(map));        for(i=0; i<t; i++)        {            scanf("%d%d",&ste[i].x,&ste[i].y);            map[ste[i].x][ste[i].y]=1;        }        for(i=0; i<t; i++)        {            sum=bfs(ste[i].x,ste[i].y);            if(sum>max1)                max1=sum;        }        printf("%d\n",max1);    }    return 0;}