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【题目链接】
【思路要点】
- 树形 算出 和 之间的最短路替换 和 之间的边。
- 此后，我们每一步或是向目标走一步，或是沿最短路走到另一棵树上。
- 直接倍增 动态 计算最短路即可。
- 时间复杂度 。
【代码】
#include<bits/stdc++.h> using namespace std; const int MAXN = 3e5 + 5; const int MAXLOG = 20; const long long INF = 1e18; typedef long long ll; typedef long double ld; typedef unsigned long long ull; template <typename T> void chkmax(T &x, T y) {x = max(x, y); } template <typename T> void chkmin(T &x, T y) {x = min(x, y); } template <typename T> void read(T &x) { x = 0; int f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -f; for (; isdigit(c); c = getchar()) x = x * 10 + c - '0'; x *= f; } template <typename T> void write(T x) { if (x < 0) x = -x, putchar('-'); if (x > 9) write(x / 10); putchar(x % 10 + '0'); } template <typename T> void writeln(T x) { write(x); puts(""); } struct state {ll dp[2][2]; }; state operator + (state a, state b) { state ans; for (int i = 0; i <= 1; i++) for (int j = 0; j <= 1; j++) ans.dp[i][j] = min(a.dp[i][0] + b.dp[0][j], a.dp[i][1] + b.dp[1][j]); return ans; } state dp[MAXN][MAXLOG], mid[MAXN]; ll val[MAXN], down[MAXN], up[MAXN]; int n, q, depth[MAXN], father[MAXN][MAXLOG]; vector <pair <int, pair <ll, ll>>> a[MAXN]; void getans(int pos, int fa, ll va, ll vb) { mid[pos].dp[0][1] = mid[pos].dp[1][0] = val[pos]; if (fa) { dp[pos][0].dp[0][0] = va; dp[pos][0].dp[1][1] = vb; dp[pos][0].dp[0][1] = INF; dp[pos][0].dp[1][0] = INF; } for (int i = 1; i < MAXLOG; i++) dp[pos][i] = dp[pos][i - 1] + mid[father[pos][i - 1]] + dp[father[pos][i - 1]][i - 1]; for (auto x : a[pos]) if (x.first != fa) getans(x.first, pos, x.second.first, x.second.second); } void dfs(int pos, int fa) { depth[pos] = depth[fa] + 1; father[pos][0] = fa; for (int i = 1; i < MAXLOG; i++) father[pos][i] = father[father[pos][i - 1]][i - 1]; down[pos] = val[pos]; for (auto x : a[pos]) if (x.first != fa) { dfs(x.first, pos); chkmin(down[pos], down[x.first] + x.second.first + x.second.second); } } void ufs(int pos, int fa) { ll Min = INF, Nin = INF; chkmin(up[pos], val[pos]); val[pos] = min(down[pos], up[pos]); for (auto x : a[pos]) if (x.first != fa) { ll tmp = down[x.first] + x.second.first + x.second.second; if (tmp < Min) { Nin = Min; Min = tmp; } else chkmin(Nin, tmp); } for (auto x : a[pos]) if (x.first != fa) { ll tmp = down[x.first] + x.second.first + x.second.second; if (tmp == Min) up[x.first] = min(Nin, up[pos]) + x.second.first + x.second.second; else up[x.first] = min(Min, up[pos]) + x.second.first + x.second.second; ufs(x.first, pos); } } int lca(int x, int y) { if (depth[x] < depth[y]) swap(x, y); for (int i = MAXLOG - 1; i >= 0; i--) if (depth[father[x][i]] >= depth[y]) x = father[x][i]; if (x == y) return x; for (int i = MAXLOG - 1; i >= 0; i--) if (father[x][i] != father[y][i]) { x = father[x][i]; y = father[y][i]; } return father[x][0]; } state unit() { state ans; ans.dp[0][0] = ans.dp[1][1] = 0; ans.dp[0][1] = ans.dp[1][0] = INF; return ans; } state gpath(int x, int y) { int z = lca(x, y), tx = x, ty = y; state ansx = unit(), ansy = unit(); for (int i = 0; i < MAXLOG; i++) { if ((depth[x] - depth[z]) & (1 << i)) { ansx = ansx + mid[tx] + dp[tx][i]; tx = father[tx][i]; } if ((depth[y] - depth[z]) & (1 << i)) { ansy = ansy + mid[ty] + dp[ty][i]; ty = father[ty][i]; } } swap(ansy.dp[0][1], ansy.dp[1][0]); return ansx + mid[z] + ansy; } int main() { read(n); for (int i = 1; i <= n; i++) read(val[i]); for (int i = 1; i <= n - 1; i++) { int x, y; ll val, vbl; read(x), read(y), read(val), read(vbl); a[x].emplace_back(y, make_pair(val, vbl)); a[y].emplace_back(x, make_pair(val, vbl)); } dfs(1, 0); up[1] = val[1]; ufs(1, 0); getans(1, 0, 0, 0); read(q); for (int i = 1; i <= q; i++) { int x, y; read(x), read(y); state tmp = gpath((x + 1) / 2, (y + 1) / 2); x = (x & 1) ^ 1, y = (y & 1) ^ 1; writeln(tmp.dp[x][y]); } return 0; }