hdu1198 Farm Irrigation —— dfs or 并查集

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1198

dfs：

#include<cstdio>//hdu1198 dfs
#include<cstring>

int well[11][5] = { {1,1,1,0,0},{1,0,1,1,0},{0,1,1,0,1},{0,0,1,1,1},{1,0,1,0,1},{0,1,1,1,0},{1,1,1,1,0},{1,1,1,0,1},{0,1,1,1,1},{1,0,1,1,1},{1,1,1,1,1}};

const int ca[5][2] = {0,1,1,0,1,1,1,2,2,1}, mov[4][2] = {1,0,-1,0,0,1,0,-1};
int m,n,sum, map[550][550],id[550][550];

void build(int set, int x, int y)
{
    for(int k = 0; k<5; k++)
    {
        if(well[set][k])
        {
            map[3*x+ca[k][0]][3*y+ca[k][1]] = 1;
        }
    }
}

void dfs(int x, int y, int iden)
{
    id[x][y] = iden;
    for(int i = 0; i<4; i++)
    {
        int xx = x + mov[i][0], yy = y + mov[i][1];
        if(xx<0 || xx>3*m-1 || yy<0 || yy>3*n-1 ) continue;
        if(map[xx][yy] && !id[xx][yy])
        dfs(xx,yy,iden);
    }
}

int main()
{
    char init[5005];
    while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1 ))
    {
        memset(map,0,sizeof(map));
        memset(id,0,sizeof(id));
        for(int i = 0; i<m; i++)
        {
            scanf("%s",init);
            for(int j = 0; j<n; j++)
            build(init[j]-65,i,j);
        }
        sum = 0;
        for(int i = 0; i<=3*m-1; i++)
        {

            for(int j = 0; j<=3*n-1; j++)
            {
                //printf("%d ",map[i][j]);
                if(map[i][j] && !id[i][j])
                {
                    dfs(i,j,++sum);
                }
            }
            //putchar('
');
        }

        printf("%d
",sum);

    }
    return 0;
}

并查集：

#include<cstdio>//hdu1198 并查集 每个格子的标号为i*n+j，初始father也为i*n+j，以次来作为合并集合的依据
#include<cstring>

int n,m,father[55][55];
char map[55][55];
char type[11][5]={"1010","1001","0110","0101","1100","0011",
               "1011","1110","0111","1101","1111"};

int find(int x)
{
    return (x==father[x/n][x%n])?x:find(father[x/n][x%n]);
}

int unit(int a, int b)
{
    a = find(a);
    b = find(b);
    if(a!=b)//合并集合
        father[a/n][a%n] = b;
}


int judge(int i, int j)
{   //判断是否与左边相连
    if(j>0 && type[map[i][j]-'A'][2]=='1' && type[map[i][j-1]-'A'][3]=='1')
        unit(i*n+j,i*n+j-1);
    //判断是否与上边相连
    if(i>0 && type[map[i][j]-'A'][0]=='1' && type[map[i-1][j]-'A'][1]=='1')
        unit(i*n+j,(i-1)*n+j);
}

int main()
{
    while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1))
    {
        for(int i = 0; i<m; i++)
        {
            scanf("%s",&map[i]);
            for(int j = 0; j<n; j++)
            {   //初始化为自己
                father[i][j] = i*n+j;
            }
        }
        
        //并查集过程
        for(int i = 0; i<m; i++)
        for(int j = 0; j<n; j++)
        judge(i,j);

        int sum = 0;
        for(int i = 0; i<m; i++)
        for(int j = 0; j<n; j++)//看看有多少个“根节点”
            if(father[i][j]==i*n+j) sum++;

        printf("%d
",sum);
    }

}