并查集应用hdu1198

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就是给并查集换了个形式

判断是否能连接那我写的比较麻烦

可以用四个字符表示四个方向

01开关

判断的时候只需要判断两个方向

一开始感觉无从想法

一步步顺着想有些还是会变简答的

#include<cstdio>#include<algorithm>using namespace std;char a[51][51];int sett[2500 + 50];int find2(int x){    while(x!=sett[x] ) x=sett[x];    return x;}void merge2(int a,int b){        int fa = find2(a);        int fb = find2(b);        if(fa > fb) sett[fa]=fb;        else sett[fb]=fa;}int main(){    int n,m;    while(scanf("%d%d",&n,&m)&&n>0&&m>0)    {        for(int i=1;i<=n;i++) scanf("%s",a[i]+1);        for(int i=1;i<=n*m;i++) sett[i]=i;//         for(int i=1;i<=n;i++){//        for(int j=1;j<=m;j++){//            printf("%d ",sett[i][j]);//        }//        printf("\n");//        }        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)        {            if(j+1<=m&& (a[i][j] == 'B'||a[i][j] == 'D'||a[i][j] == 'F'||a[i][j] == 'G'||a[i][j] == 'I'||a[i][j] == 'J'||a[i][j] == 'K') && (a[i][j+1]=='A'||a[i][j+1]=='C'||a[i][j+1]=='F'||a[i][j+1]=='G'||a[i][j+1]=='H'||a[i][j+1]=='I'||a[i][j+1]=='K'))                merge2(sett[(i-1)*m+j],sett[(i-1)*m+j+1]);            if(i+1<=n&& (a[i][j] == 'C'||a[i][j] == 'D'||a[i][j] == 'E'||a[i][j] == 'H'||a[i][j] == 'I'||a[i][j] == 'J'||a[i][j] == 'K') && (a[i+1][j]=='A'||a[i+1][j]=='B'||a[i+1][j]=='E'||a[i+1][j]=='G'||a[i+1][j]=='H'||a[i+1][j]=='J'||a[i+1][j]=='K'))                 merge2(sett[(i-1)*m+j],sett[(i)*m+j]);        }//        for(int i=1;i<=n*m;i++) printf("%d ",i);printf("\n");        int countt=0;        for(int i=1;i<=n*m;i++) if(sett[i] == i) countt++;        printf("%d\n",countt);    }    return 0;}

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