hdu1198 Farm Irrigation —— dfs or 并查集_一张完全图断掉一些边问联通块数量-优快云博客

本文提供了一道名为 HDU 1198 的编程题解，使用两种不同的算法解决：深度优先搜索 (DFS) 和并查集。通过这两种方法详细解释了如何在给定的地图上寻找连通区域的数量。

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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('\n');
        }

        printf("%d\n",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\n",sum);
    }

}