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poj1164 The Castle 水搜索
http://poj.org/problem?id=1164
The Castle
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 6382 | Accepted: 3610 |
Description
1 2 3 4 5 6 7 ############################# 1 # | # | # | | # #####---#####---#---#####---# 2 # # | # # # # # #---#####---#####---#####---# 3 # | | # # # # # #---#########---#####---#---# 4 # # | | | | # # ############################# (Figure 1) # = Wall | = No wall - = No wall
Figure 1 shows the map of a castle.Write a program that calculates
1. how many rooms the castle has
2. how big the largest room is
The castle is divided into m * n (m<=50, n<=50) square modules. Each such module can have between zero and four walls.
Input
Your program is to read from standard input. The first line contains the number of modules in the north-south direction and the number of modules in the east-west direction. In the following lines each module is described by a number (0 <= p <= 15). This number is the sum of: 1 (= wall to the west), 2 (= wall to the north), 4 (= wall to the east), 8 (= wall to the south). Inner walls are defined twice; a wall to the south in module 1,1 is also indicated as a wall to the north in module 2,1. The castle always has at least two rooms.
Output
Your program is to write to standard output: First the number of rooms, then the area of the largest room (counted in modules).
Sample Input
4 7 11 6 11 6 3 10 6 7 9 6 13 5 15 5 1 10 12 7 13 7 5 13 11 10 8 10 12 13
Sample Output
5 9
Source
IOI 1994
水题。就是m*n的房子每个点四周可能有墙，1 2 4 8分别代表哪四个方向有墙。告诉每个格子四周之和，问连通块个数和最大连通快个数。
分析：首先由每个格子的数可以唯一确定这个点四个方向是否有墙(很有意思由2的n次方可以唯一组成2^(n+1)以下所有数)，具体的判断方法是通过位运算!!这样我们就dfs或者bfs找连通块了。。
/**
* @author neko01
*/
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=55;
int a[N][N];
bool vis[N][N];
int ans,num;
int n,m;
int dir[4][2]={0,-1,-1,0,0,1,1,0};
bool immap(int x,int y)
{
return x>=0&&y>=0&&x<n&&y<m;
}
void bfs(int sx,int sy)
{
vis[sx][sy]=true;
queue<pp>q;
q.push(mp(sx,sy));
int tmp=0;
while(!q.empty())
{
pp cur=q.front();
q.pop();
int x=cur.first;
int y=cur.second;
tmp++;
for(int i=0;i<4;i++)
{
int xx=x+dir[i][0];
int yy=y+dir[i][1];
if(!((1<<i)&a[x][y])&&immap(xx,yy)&&!vis[xx][yy])
{
q.push(mp(xx,yy));
vis[xx][yy]=true;
}
}
}
if(tmp>ans) ans=tmp;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
scanf("%d",&a[i][j]);
clr(vis);
ans=num=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(!vis[i][j])
{
bfs(i,j);
num++;
}
}
printf("%d\n%d\n",num,ans);
}
return 0;
}poj1164 The Castle 水搜索
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