洛谷上的lca模板题——传送门
1.tarjan求lca
学了求lca的tarjan算法（离线），在洛谷上做模板题，结果后三个点超时。
又把询问改成链式前向星，才ok。
这个博客，tarjan分析的很详细。
附代码——
#include <cstdio> #include <cstring> const int maxn = 500001; int n, m, cnt, s, cns; int x, y, z[maxn];//z是x和y的lca int f[maxn], head[maxn], from[maxn]; bool vis[maxn]; struct node { int to, next; }e[2 * maxn]; struct Node { int to, next, num; }q[2 * maxn]; inline int read()//读入优化 { int x = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x * f; } inline void ask(int u, int v, int i)//储存待询问的结构体，也是链式前向星优化 { q[cns].num = i;//num表示第几次询问 q[cns].to = v; q[cns].next = from[u]; from[u] = cns++; } inline void add(int u, int v)// { e[cnt].to = v; e[cnt].next = head[u]; head[u] = cnt++; } inline int find(int a) { return a == f[a] ? a : f[a] = find(f[a]);//路径压缩优化 } /*inline void Union(int a, int b) { int fx = find(a), fy = find(b); if(fx == fy) return; f[fy] = fx; }*/ inline void tarjan(int k) { int i, j; vis[k] = 1; f[k] = k; for(i = head[k]; i != -1; i = e[i].next) if(!vis[e[i].to]) { tarjan(e[i].to); //Union(k, e[i].to); f[e[i].to] = k; } for(i = from[k]; i != -1; i = q[i].next) if(vis[q[i].to] == 1) z[q[i].num] = find(q[i].to); } int main() { int i, j, u, v; n = read(); m = read(); s = read(); memset(head, -1, sizeof(head)); memset(from, -1, sizeof(from)); for(i = 1; i <= n - 1; i++) { u = read(); v = read(); add(u, v);//注意添加两遍 add(v, u); } for(i = 1; i <= m; i++) { x = read(); y = read(); ask(x, y, i);//两遍 ask(y, x, i); } tarjan(s); for(i = 1; i <= m; i++) printf("%d ", z[i]); return 0; }
进过培训，修改了代码
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <string> 5 # include <cmath> 6 # include <vector> 7 # include <map> 8 # include <queue> 9 # include <cstdlib> 10 # define MAXN 500001 11 using namespace std; 12 13 inline int get_num() { 14 int k = 0, f = 1; 15 char c = getchar(); 16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1; 17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0'; 18 return k * f; 19 } 20 21 int n, m, s; 22 int fa[MAXN], qx[MAXN], qy[MAXN], ans[MAXN], f[MAXN]; 23 vector <int> vec[MAXN], q[MAXN]; 24 25 inline int find(int x) 26 { 27 return x == fa[x] ? x : fa[x] = find(fa[x]); 28 } 29 30 inline void dfs(int u) 31 { 32 int i, v; 33 fa[u] = u; 34 for(i = 0; i < vec[u].size(); i++) 35 { 36 v = vec[u][i]; 37 if(f[u] != v) f[v] = u, dfs(v); 38 } 39 for(i = 0; i < q[u].size(); i++) 40 if(f[v = u ^ qx[q[u][i]] ^ qy[q[u][i]]]) 41 ans[q[u][i]] = find(v); 42 fa[u] = f[u]; 43 } 44 45 int main() 46 { 47 int i, x, y; 48 n = get_num(); 49 m = get_num(); 50 s = get_num(); 51 for(i = 1; i < n; i++) 52 { 53 x = get_num(); 54 y = get_num(); 55 vec[x].push_back(y); 56 vec[y].push_back(x); 57 } 58 for(i = 1; i <= m; i++) 59 { 60 qx[i] = get_num(); 61 qy[i] = get_num(); 62 q[qx[i]].push_back(i); 63 q[qy[i]].push_back(i); 64 } 65 dfs(s); 66 for(i = 1; i <= m; i++) printf("%d ", ans[i]); 67 return 0; 68 }
其实上面两个代码有些重复运算，请手动把求lca的过程放到dfs上面（也就是遍历到这个节点就求lca，而不是遍历完再求）
2.倍增求lca
下面是求lca的倍增算法（在线）。
1. DFS预处理出所有节点的深度和父节点
inline void dfs(int u) { int i; for(i=head[u];i!=-1;i=next[i]) { if (!deep[to[i]]) { deep[to[i]] = deep[u]+1; p[to[i]][0] = u; //p[x][0]保存x的父节点为u; dfs(to[i]); } } }
2. 初始各个点的2^j祖先是谁 ,其中 2^j (j =0...log(该点深度))倍祖先，1倍祖先就是父亲，2倍祖先是父亲的父亲......。
void init() { int i,j; //p[i][j]表示i结点的第2^j祖先 for(j=1;(1<<j)<=n;j++) for(i=1;i<=n;i++) if(p[i][j-1]!=-1) p[i][j]=p[p[i][j-1]][j-1];//i的第2^j祖先就是i的第2^(j-1)祖先的第2^(j-1)祖先 }
3.从深度大的节点上升至深度小的节点同层，如果此时两节点相同直接返回此节点，即lca。
否则，利用倍增法找到最小深度的 p[a][j]!=p[b][j]，此时他们的父亲p[a][0]即lca。
int lca(int a,int b)//最近公共祖先 { int i,j; if(deep[a]<deep[b])swap(a,b); for(i=0;(1<<i)<=deep[a];i++); i--; //使a，b两点的深度相同 for(j=i;j>=0;j--) if(deep[a]-(1<<j)>=deep[b]) a=p[a][j]; if(a==b)return a; //倍增法，每次向上进深度2^j，找到最近公共祖先的子结点 for(j=i;j>=0;j--) { if(p[a][j]!=-1&&p[a][j]!=p[b][j]) { a=p[a][j]; b=p[b][j]; } } return p[a][0]; }
最后是完整代码，为了节约时间，就没有把p数组初始化为-1.
#include <cstdio> #include <cstring> #include <iostream> const int maxn = 500001; int n, m, cnt, s; int next[2 * maxn], to[2 * maxn], head[2 * maxn], deep[maxn], p[maxn][21]; inline void add(int x, int y) { to[cnt] = y; next[cnt] = head[x]; head[x] = cnt++; } inline void dfs(int i) { int j; for(j = head[i]; j != -1; j = next[j]) if(!deep[to[j]]) { deep[to[j]] = deep[i] + 1; p[to[j]][0] = i; dfs(to[j]); } } inline void init() { int i, j; for(j = 1; (1 << j) <= n; j++) for(i = 1; i <= n; i++) p[i][j] = p[p[i][j - 1]][j - 1]; } inline int lca(int a, int b) { int i, j; if(deep[a] < deep[b]) std::swap(a, b); for(i = 0; (1 << i) <= deep[a]; i++); i--; for(j = i; j >= 0; j--) if(deep[a] - (1 << j) >= deep[b]) a = p[a][j]; if(a == b) return a; for(j = i; j >= 0; j--) if(p[a][j] != p[b][j]) { a = p[a][j]; b = p[b][j]; } return p[a][0]; } int main() { int i, j, x, y; memset(head, -1, sizeof(head)); scanf("%d %d %d", &n, &m, &s); for(i = 1; i <= n - 1; i++) { scanf("%d %d", &x, &y); add(x, y); add(y, x); } deep[s] = 1; dfs(s); init(); for(i = 1; i <= m; i++) { scanf("%d %d", &x, &y); printf("%d ", lca(x, y)); } return 0; }
经过培训，又改了改代码。
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <string> 5 # include <cmath> 6 # include <vector> 7 # include <map> 8 # include <queue> 9 # include <cstdlib> 10 # define MAXN 500001 11 using namespace std; 12 13 inline int get_num() { 14 int k = 0, f = 1; 15 char c = getchar(); 16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1; 17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0'; 18 return k * f; 19 } 20 21 int n, m, s; 22 int f[MAXN][25], deep[MAXN]; 23 vector <int> vec[MAXN]; 24 25 inline void dfs(int u) 26 { 27 int i, v; 28 deep[u] = deep[f[u][0]] + 1; 29 for(i = 0; f[u][i]; i++) f[u][i + 1] = f[f[u][i]][i]; 30 for(i = 0; i < vec[u].size(); i++) 31 { 32 v = vec[u][i]; 33 if(!deep[v]) f[v][0] = u, dfs(v); 34 } 35 } 36 37 inline int lca(int x, int y) 38 { 39 int i; 40 if(deep[x] < deep[y]) swap(x, y); 41 for(i = 20; i >= 0; i--) 42 if(deep[f[x][i]] >= deep[y]) 43 x = f[x][i]; 44 if(x == y) return x; 45 for(i = 20; i >= 0; i--) 46 if(f[x][i] != f[y][i]) 47 x = f[x][i], y = f[y][i]; 48 return f[x][0]; 49 } 50 51 int main() 52 { 53 int i, x, y; 54 n = get_num(); 55 m = get_num(); 56 s = get_num(); 57 for(i = 1; i < n; i++) 58 { 59 x = get_num(); 60 y = get_num(); 61 vec[x].push_back(y); 62 vec[y].push_back(x); 63 } 64 dfs(s); 65 for(i = 1; i <= m; i++) 66 { 67 scanf("%d %d", &x, &y); 68 printf("%d ", lca(x, y)); 69 } 70 return 0; 71 }
3.树剖法求lca
1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <string> 5 # include <cmath> 6 # include <vector> 7 # include <map> 8 # include <queue> 9 # include <cstdlib> 10 # define MAXN 500001 11 using namespace std; 12 13 inline int get_num() { 14 int k = 0, f = 1; 15 char c = getchar(); 16 for(; !isdigit(c); c = getchar()) if(c == '-') f = -1; 17 for(; isdigit(c); c = getchar()) k = k * 10 + c - '0'; 18 return k * f; 19 } 20 21 int n, m, s; 22 int f[MAXN], size[MAXN], top[MAXN], son[MAXN], deep[MAXN]; 23 vector <int> vec[MAXN]; 24 25 inline void dfs1(int u) 26 { 27 int i, v; 28 size[u] = 1; 29 deep[u] = deep[f[u]] + 1; 30 for(i = 0; i < vec[u].size(); i++) 31 { 32 v = vec[u][i]; 33 if(!deep[v]) 34 { 35 f[v] = u; 36 dfs1(v); 37 size[u] += size[v]; 38 if(size[son[u]] < size[v]) son[u] = v; 39 } 40 } 41 } 42 43 inline void dfs2(int u, int tp) 44 { 45 int i, v; 46 top[u] = tp; 47 if(!son[u]) return; 48 dfs2(son[u], tp); 49 for(i = 0; i < vec[u].size(); i++) 50 { 51 v = vec[u][i]; 52 if(v != son[u] && v != f[u]) dfs2(v, v); 53 } 54 } 55 56 inline int lca(int x, int y) 57 { 58 while(top[x] != top[y]) 59 { 60 if(deep[top[x]] < deep[top[y]]) swap(x, y); 61 x = f[top[x]]; 62 } 63 if(deep[x] > deep[y]) swap(x, y); 64 return x; 65 } 66 67 int main() 68 { 69 int i, x, y; 70 n = get_num(); 71 m = get_num(); 72 s = get_num(); 73 for(i = 1; i < n; i++) 74 { 75 x = get_num(); 76 y = get_num(); 77 vec[x].push_back(y); 78 vec[y].push_back(x); 79 } 80 dfs1(s); 81 dfs2(s, s); 82 for(i = 1; i <= m; i++) 83 { 84 x = get_num(); 85 y = get_num(); 86 printf("%d ", lca(x, y)); 87 } 88 return 0; 89 }