hdu 1198(并查集 )

自从懂了并查集只后，感觉好多题都是并查集，就像哪一天的字典树一样，这道题一看就是一个并查集，最后查询父节点有几个，

难点：建模的时候应该吧上下联通的和左右联通的标记一下，只要他们和上下左右的都能连通，就把他们并到一个集合里面，我是只判断下和右即可，

源代码：

#include
#include
int up[8], down[8], right[8], left[8];
int par[2504];
char map[54][54];
char mp1[100][100];
char mp2[100][100];
void init()
{
    up[0] = 'A'; up[1] = 'B'; up[2] = 'E'; up[3] = 'G'; up[4] = 'H'; up[5] = 'J'; up[6] = 'K';
    down[0] = 'C'; down[1] = 'D'; down[2] = 'E'; down[3] = 'H'; down[4] = 'I'; down[5] = 'J'; down[6] = 'K';
    left[0] = 'A'; left[1] = 'C'; left[2] = 'F'; left[3] = 'G'; left[4] = 'H'; left[5] = 'I'; left[6] = 'K';
    right[0] = 'B'; right[1] = 'D'; right[2] = 'F'; right[3] = 'G'; right[4] = 'I'; right[5] = 'J'; right[6] = 'K';
    memset(mp1, 0, sizeof(mp1));
    memset(mp2, 0, sizeof(mp2));
    for (int i = 0; i <= 6; i++){
        for (int j = 0; j <= 6; j++){
            mp1[down[i]][up[j]] = 1;
            mp2[right[i]][left[j]] = 1;
        }
    }
}

int findset(int x){
   if(x!=par[x])
   par[x]=findset(par[x]);
   return par[x];
}

void unite(int x,int y){
   int px=findset(x);
   int py=findset(y);
   if(px!=py)
   par[py]=px;
}
int main()
{
    int M, N;
    init();
    while (scanf("%d%d", &M, &N) != EOF){
        if (M == -1 && N == -1) break;
        for (int i = 0; i < M; i++){
            scanf("%s", map[i]);
        }

        for (int i = 0; i <= M * N; i++){
            par[i] = i;
        }
        for (int i = 0; i < M; i++){
            for (int j = 0; j < N; j++){
                if (j < N - 1 && mp2[map[i][j]][map[i][j+1]] == 1){
                    int t1 = i * N + j;
                    int t2 = i * N + j + 1;
                    unite(t1, t2);
                }

                if (i < M - 1 && mp1[map[i][j]][map[i + 1][j]] == 1){
                    int t1 = i * N + j;
                    int t2 = (i + 1) * N + j;
                    unite(t1, t2);
                }
            }
        }

        int ans = 0;
        for (int i = 0; i < M * N; i++){
            if (par[i] == i){
                ans++;
            }
        }

        printf("%d\n", ans);
    }
    return 0;
}