前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >Farm Irrigation

Farm Irrigation

作者头像
用户1624346
发布2018-04-17 15:47:12
4160
发布2018-04-17 15:47:12
举报
文章被收录于专栏:calmoundcalmoundcalmound

题意:判断图中的图形有几个集合。

        这道题做的郁闷死我了,为了找开始字母的错误一个一个匹对。

        这道题让我也懂得了表格的位置可以代表并查集的点,这样每个都不一样,而我想的那种用字母是否本身来判断有多少个集合是错误的,很明显一个字母不会只出现一次。而且在计算位置的时候也出错了,应该i*列代表的第几行

#include<stdio.h>
#include<string.h>
const int MAXN=110;
int total;
int gra_row[MAXN][MAXN],gra_col[MAXN][MAXN];
int rank[2510],father[2510];

void init()
{
   memset(gra_row,0,sizeof(gra_row));
   memset(gra_col,0,sizeof(gra_col));
   gra_row['B']['A']=gra_row['B']['C']=gra_row['B']['F']=gra_row['B']['G']=gra_row['B']['H']=gra_row['B']['I']=gra_row['B']['K']=1;
   gra_row['D']['A']=gra_row['D']['C']=gra_row['D']['F']=gra_row['D']['G']=gra_row['D']['H']=gra_row['D']['I']=gra_row['D']['K']=1;
   gra_row['F']['A']=gra_row['F']['C']=gra_row['F']['F']=gra_row['F']['G']=gra_row['F']['H']=gra_row['F']['I']=gra_row['F']['K']=1;
   gra_row['G']['A']=gra_row['G']['C']=gra_row['G']['F']=gra_row['G']['G']=gra_row['G']['H']=gra_row['G']['I']=gra_row['G']['K']=1;
   gra_row['I']['A']=gra_row['I']['C']=gra_row['I']['F']=gra_row['I']['G']=gra_row['I']['H']=gra_row['I']['I']=gra_row['I']['K']=1;
   gra_row['J']['A']=gra_row['J']['C']=gra_row['J']['F']=gra_row['J']['G']=gra_row['J']['H']=gra_row['J']['I']=gra_row['J']['K']=1;
   gra_row['K']['A']=gra_row['K']['C']=gra_row['K']['F']=gra_row['K']['G']=gra_row['K']['H']=gra_row['K']['I']=gra_row['K']['K']=1;



   gra_col['C']['A']=gra_col['C']['B']=gra_col['C']['E']=gra_col['C']['G']=gra_col['C']['H']=gra_col['C']['J']=gra_col['C']['K']=1;
   gra_col['D']['A']=gra_col['D']['B']=gra_col['D']['E']=gra_col['D']['G']=gra_col['D']['H']=gra_col['D']['J']=gra_col['D']['K']=1;
   gra_col['E']['A']=gra_col['E']['B']=gra_col['E']['E']=gra_col['E']['G']=gra_col['E']['H']=gra_col['E']['J']=gra_col['E']['K']=1;
   gra_col['H']['A']=gra_col['H']['B']=gra_col['H']['E']=gra_col['H']['G']=gra_col['H']['H']=gra_col['H']['J']=gra_col['H']['K']=1;
   gra_col['I']['A']=gra_col['I']['B']=gra_col['I']['E']=gra_col['I']['G']=gra_col['I']['H']=gra_col['I']['J']=gra_col['I']['K']=1;
   gra_col['J']['A']=gra_col['J']['B']=gra_col['J']['E']=gra_col['J']['G']=gra_col['J']['H']=gra_col['J']['J']=gra_col['J']['K']=1;
   gra_col['K']['A']=gra_col['K']['B']=gra_col['K']['E']=gra_col['K']['G']=gra_col['K']['H']=gra_col['K']['J']=gra_col['K']['K']=1;

}

void Make_set()
{
    for(int i=0;i<=2505;i++)
    {
        father[i]=i;
        rank[i]=0;
    }
}

int Find(int x)
{
    if(x!=father[x])
    {
        return father[x]=Find(father[x]);
    }
    return x;
}
void Union(int x,int y)
{
    x=Find(x);
    y=Find(y);
    if(x==y) return ;
    total--;
    if(rank[x]>rank[y]) father[y]=x;
    else
    {
        if(rank[x]==rank[y])
        {
            rank[y]++;
        }
        father[x]=y;
    }
}
int main()
{
    int n,m,i,j;
    int a,b;
    char str[MAXN][MAXN];
    init();
    while(scanf("%d%d",&n,&m))
    {
        Make_set();
        memset(str,0,sizeof(str));
        total=m*n;
        if(m<0 || n<0) break;
        for(i=0;i<n;i++)
        {
            scanf("%s",str[i]);
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(gra_row[str[i][j]][str[i][j+1]]==1) //Union(str[i][j],str[i][j+1]);
                {
                    a=i*m+j;
                    b=a+1;
                    Union(a,b);
                }
                if(gra_col[str[i][j]][str[i+1][j]]==1) //Union(str[i][j],str[i+1][j]);
                {
                    a=i*m+j;
                    b=(i+1)*m+j;
                    Union(a,b);
                }
            }
        }
        printf("%d\n",total);
    }
    return 0;
}

简便一点的写法

char up[8]={"ABEGHJK"},
        down[8]={"CDEHIJK"},
        left[8]={"ACFGHIK"},
        right[8]={"BDFGIJK"};
int row_link[30][30]={0},column_link[30][30]={0};
void init()
{
    int i,j;
    for(i=0;i<7;i++)
    {
         for(j=0;j<7;j++)
        {
            column_link[down[i]-'A'][up[j]-'A']=1;
            row_link[right[i]-'A'][left[j]-'A']=1;
        }
    }
}
本文参与 腾讯云自媒体分享计划,分享自作者个人站点/博客。
原始发表:2012-08-12 ,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体分享计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档