1.开两个并查集 f[x] 表示 x 的同类 f[x + n] 表示 x 的敌人
——代码
1 #include <cstdio> 2 #include <iostream> 3 #define N 200001 4 5 int T, n, m; 6 int f[N]; 7 8 inline int read() 9 { 10 int x = 0, f = 1; 11 char ch = getchar(); 12 for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; 13 for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; 14 return x * f; 15 } 16 17 inline int find(int x) 18 { 19 return x == f[x] ? x : f[x] = find(f[x]); 20 } 21 22 inline void connect(int x, int y) 23 { 24 x = find(x); 25 y = find(y); 26 f[x] = y; 27 } 28 29 int main() 30 { 31 int i, j, x, y; 32 char s[1]; 33 T = read(); 34 while(T--) 35 { 36 n = read(); 37 m = read(); 38 for(i = 1; i <= (n << 1); i++) f[i] = i; 39 for(i = 1; i <= m; i++) 40 { 41 scanf("%s", s); 42 x = read(); 43 y = read(); 44 if(s[0] == 'A') 45 { 46 if(find(x) == find(y)) puts("In the same gang."); 47 else if(find(x) == find(y + n) || find(y) == find(x + n)) puts("In different gangs."); 48 else puts("Not sure yet."); 49 } 50 else 51 { 52 connect(x, y + n); 53 connect(y, x + n); 54 } 55 } 56 } 57 }
2.带权并查集
——代码
1 #include <cstdio> 2 #include <iostream> 3 #define N 1000001 4 5 int T, n, m; 6 int f[N], d[N]; 7 8 inline int read() 9 { 10 int x = 0, f = 1; 11 char ch = getchar(); 12 for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; 13 for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; 14 return x * f; 15 } 16 17 inline int find(int x) 18 { 19 if(x ^ f[x]) 20 { 21 int fx = f[x]; 22 f[x] = find(f[x]); 23 d[x] = (d[x] + d[fx]) % 2; 24 } 25 return f[x]; 26 } 27 28 inline void connect(int x, int y) 29 { 30 int fx = find(x), fy = find(y); 31 d[fx] = (d[y] + 1 - d[x]) % 2; 32 f[fx] = f[y]; 33 } 34 35 int main() 36 { 37 int i, x, y; 38 char s[1]; 39 T = read(); 40 while(T--) 41 { 42 n = read(); 43 m = read(); 44 for(i = 1; i <= n; i++) f[i] = i, d[i] = 0; 45 for(i = 1; i <= m; i++) 46 { 47 scanf("%s", s); 48 x = read(); 49 y = read(); 50 if(s[0] == 'D') connect(x, y); 51 else 52 { 53 if(find(x) ^ find(y)) puts("Not sure yet."); 54 else if(d[x] == d[y]) puts("In the same gang."); 55 else puts("In different gangs."); 56 } 57 } 58 } 59 return 0; 60 }