vector<int> tree[MAX]; int dep[MAX]; int fas[MAX];
namespace M1 { voiddfs(int x, int fa){ if (fa != -1) { dep[x] = dep[fa] + 1; } fas[x] = fa; for (int i = 0; i < tree[x].size(); i++) { if (fas[tree[x].at(i)] == -2) M1::dfs(tree[x].at(i), x); } }
intsolve(int a, int b){ while (1) { if (a == b) { return a; } elseif (fas[a] == fas[b]) { return fas[a]; } elseif (fas[b] == a) { return a; } elseif (fas[a] == b) { return b; } int da = dep[a], db = dep[b]; int delta = abs(da - db); if (da > db) { for (int i = 0; i < delta; i++) { a = fas[a]; da = dep[a]; } } elseif (da < db) { for (int i = 0; i < delta; i++) { b = fas[b]; db = dep[b]; } } else { a = fas[a]; da = dep[a]; } } return-1; } }
bool first = true;
intLCA(int a, int b, int r){ if (first) { M1::dfs(r, -1); first = false; } int res; res = M1::solve(a, b); return res; }
intmain(int argc, char *argv[]){ int m, n, s; cin >> m >> n >> s; for (int i = 0; i < MAX; i++) { dep[i] = 0; fas[i] = -2; } for (int i = 1; i < n; i++) { int x, y; cin >> x >> y; tree[x].push_back(y); tree[y].push_back(x); } dep[s] = 0; fas[s] = -1; for (int i = 0; i < m; i++) { int a, b; cin >> a >> b; int res = LCA(a, b, s); cout << res << endl; } return0; }
注意這道題目的資料輸入,x y表示x 結點和 y 結點之間有一條直接連線的邊(資料保證可以構成樹)。 所以需要用鄰接表的形式,表示多叉樹。
vector<int> tree[MAX]; int dep[MAX]; int fas[MAX][MUL_MAX]; bool first = true;
intlg2(int x){ returnlog(x) / log(2) + 1; }
voiddfs(int x, int fa){ if (fa != -1) { dep[x] = dep[fa] + 1; } fas[x][0] = fa; for (int i = 1; (1 << i) <= dep[x]; i++) { fas[x][i] = fas[fas[x][i - 1]][i - 1]; } for (int i = 0; i < tree[x].size(); i++) { if (tree[x].at(i) != fa) dfs(tree[x].at(i), x); } }
intsolve(int a, int b){ if (dep[b] > dep[a]) swap(a, b); while (dep[a] > dep[b]) { a = fas[a][lg2(dep[a] - dep[b]) - 1]; } if (a == b) return a; for (int i = lg2(dep[a]); i >= 0; i--) { if (fas[a][i] != fas[b][i]) { a = fas[a][i]; b = fas[b][i]; } } return fas[a][0]; }
intLCA(int a, int b, int r){ if (first) { dep[r] = 0; fas[r][0] = -1; dfs(r, -1); first = false; } int res; res = solve(a, b); return res; }
intmain(int argc, char *argv[]){ int m, n, s; cin >> m >> n >> s; for (int i = 0; i < MAX; i++) { dep[i] = 0; for (int j = 0; j < MUL_MAX; j++) fas[i][j] = -2; } for (int i = 1; i < n; i++) { int x, y; cin >> x >> y; tree[x].push_back(y); tree[y].push_back(x); } for (int i = 0; i < m; i++) { int a, b; cin >> a >> b; int res = LCA(a, b, s); cout << res << endl; } return0; }