HDU 1198 Farm Irrigation (并查集)

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5809 Accepted Submission(s): 2516


[align=left]Problem Description[/align]
Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int MAXN=55;
char ch;
int a[MAXN][MAXN],p[MAXN*MAXN];
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};//寻找方向:上右下左
int map[11][4]={{1,0,0,1},//从上右下左四个方向依次列出每一个方块的连通情况
{1,1,0,0},
{0,0,1,1},
{0,1,1,0},
{1,0,1,0},
{0,1,0,1},
{1,1,0,1},
{1,0,1,1},
{0,1,1,1},
{1,1,1,0},
{1,1,1,1}
};
int Find(int x)
{
return p[x]==x?x:p[x]=Find(p[x]);
}

void Union(int a,int b)
{
int x=Find(a);
int y=Find(b);
if(x!=y)
p[x]=y;
}

int main()
{
//freopen("in.txt","r",stdin);
int n,m;
while(scanf("%d %d%*c",&m,&n)&&m>0)
{
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
scanf("%c",&ch);
a[i][j]=ch-'A';
}
getchar();
}

for(int i=0;i<n*m;i++)
p[i]=i;

for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
for(int k=0;k<4;k++)
{
int next_x,next_y;
next_x=i+dir[k][0];
next_y=j+dir[k][1];
if(0<=next_x&&next_x<m&&0<=next_y&&next_y<n)
{
if(k==0)
{
if(map[a[next_x][next_y]][2]&&map[a[i][j]][0])
Union(next_x*n+next_y,i*n+j);
}
if(k==1)
{
if(map[a[next_x][next_y]][3]&&map[a[i][j]][1])
Union(next_x*n+next_y,i*n+j);
}
if(k==2)
{
if(map[a[next_x][next_y]][0]&&map[a[i][j]][2])
Union(next_x*n+next_y,i*n+j);
}
if(k==3)
{
if(map[a[next_x][next_y]][1]&&map[a[i][j]][3])
Union(next_x*n+next_y,i*n+j);
}
}
}
}
}
//        for(int i=0;i<n*m;i++)
//            printf("%d ",p[i]);
//        printf("\n");

int cnt=0;
for(int i=0;i<n*m;i++)
if(p[i]==i)
cnt++;
printf("%d\n",cnt);
}
}

View Code