CF1000G. Two-Paths

Solution

我們發現除了樹上$(x,y)$最短路徑上的邊經過一次，其余邊要么走$0$次，要么走$2$次。

因此考慮先假設每條邊走兩次，最后把走一次的邊的貢獻加上。

我們把從$(x,y)$路徑上的點擴展出去連痛塊的貢獻拆成三部分：

$x,y$的子樹中的部分。

$x,y$路徑上不是端點的點延伸出去且不經過$(x,y)$路徑的部分。

$x,y$的$lca$向上延伸的部分。

我們可以通過樹形$dp$來求出這三種貢獻，最后詢問就形如路徑求和，倍增或者樹剖即可。

時間復雜度$O(n\log n)$。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y <= x ? x = y, 1 : 0; } template<typename T> inline bool upmax(T &x, T y) { return x <= y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se secondtypedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int, int> PR; typedef vector<int> VI; const lod eps = 1e-9; const lod pi = acos(-1); const int oo = 1 << 30; const ll loo = (1ll << 62) - 1; const int MAXN = 600005; const int mods = 998244353; const int MX = 100000; const int inv2 = (mods + 1) >> 1; const int INF = 0x3f3f3f3f; //1061109567 /*--------------------------------------------------------------------*/namespace FastIO{constexpr int SIZE = (1 << 21) + 1;int num = 0, f;char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)inline void flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}inline void putc(char c) {*oS ++ = c;if (oS == oT) flush();}inline void getc(char &c) {for (c = gc(); !isalpha(c) && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); isalpha(c) ; c = gc()) st[++ n] = c;st[++ n] = '\0';}template<class I>inline void read(I &x) {for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);x *= f;}template<class I>inline void print(I x) {if (x < 0) putc('-'), x = -x;if (!x) putc('0');while (x) que[++ num] = x % 10 + 48, x /= 10;while (num) putc(que[num --]);}inline void putstr(string st) {for (int i = 0; i < (int)st.size() ; ++ i) putc(st[i]);}struct Flusher_{~Flusher_(){flush();}} io_Flusher_; } using FastIO :: read; using FastIO :: putc; using FastIO :: putstr; using FastIO :: reads; using FastIO :: print;struct enode{ int nxt, to, c; } e[MAXN << 1]; int head[MAXN], C[MAXN], fa[MAXN][20], dep[MAXN], a[MAXN], Log[MAXN], edgenum = 0; ll f[MAXN], g[MAXN], h[MAXN], sum[MAXN][20];void add(int u, int v, int c) {e[++ edgenum] = (enode){head[u], v, c}, head[u] = edgenum; } void tree_dp(int x, int father) {for (int i = head[x]; i ; i = e[i].nxt) {int v = e[i].to;if (v == father) continue;tree_dp(v, x);}f[x] = a[x];for (int i = head[x]; i ; i = e[i].nxt) {int v = e[i].to, c = e[i].c;if (v == father) continue;f[x] += max(f[v] - c - c, 0ll);}for (int i = head[x]; i ; i = e[i].nxt) {int v = e[i].to, c = e[i].c;if (v == father) continue;g[v] = f[x] - max(f[v] - c - c, 0ll);} } void dfs(int x, int father) {dep[x] = dep[father] + 1, fa[x][0] = father, sum[x][0] = g[x] - C[x];for (int i = 1; i <= Log[dep[x]] ; ++ i) fa[x][i] = fa[fa[x][i - 1]][i - 1], sum[x][i] = sum[x][i - 1] + sum[fa[x][i - 1]][i - 1];for (int i = head[x]; i ; i = e[i].nxt) {int v = e[i].to, c = e[i].c;if (v == father) continue;h[v] = max(h[x] + f[x] - max(f[v] - c - c, 0ll) - c - c, 0ll);C[v] = c;dfs(v, x);} } int getlca(int x, int y) {if (dep[x] < dep[y]) swap(x, y);for (int i = Log[dep[x]]; i >= 0 ; -- i)if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];if (x == y) return x;for (int i = Log[dep[x]]; i >= 0 ; -- i)if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];return fa[x][0]; } int jump(int u, int d) {for (int i = Log[d]; i >= 0 ; -- i)if ((d >> i) & 1) u = fa[u][i];return u; } ll getsum(int u, int d) {ll ans = 0;for (int i = Log[d]; i >= 0 ; -- i)if ((d >> i) & 1) ans += sum[u][i], u = fa[u][i];return ans; }signed main() { #ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin); #endif int n, Case;read(n), read(Case);for (int i = 1; i <= n ; ++ i) read(a[i]);for (int i = 1, u, v, c; i < n ; ++ i) read(u), read(v), read(c), add(u, v, c), add(v, u, c);tree_dp(1, 0);dep[0] = -1, Log[1] = 0;for (int i = 2; i <= n ; ++ i) Log[i] = Log[i >> 1] + 1;dfs(1, 0);// for (int i = 1; i <= n ; ++ i) cout << i << ":" << f[i] << " " << g[i] << " " << h[i] << endl;while (Case --) {int u, v;read(u), read(v);if (dep[u] < dep[v]) swap(u, v);ll ans = 0;int t = getlca(u, v);if (u == v) ans = h[u] + f[u];else if (v == t) {int uu = jump(u, dep[u] - dep[t] - 1);ans = h[t] + f[t] - max(f[uu] - C[uu] - C[uu], 0ll) + getsum(u, dep[u] - dep[t] - 1) + f[u] - C[uu];}else {int uu = jump(u, dep[u] - dep[t] - 1), vv = jump(v, dep[v] - dep[t] - 1);ans = h[t] + f[t] - max(f[uu] - C[uu] - C[uu], 0ll) - max(f[vv] - C[vv] - C[vv], 0ll) - C[uu] - C[vv];ans += getsum(u, dep[u] - dep[t] - 1) + getsum(v, dep[v] - dep[t] - 1) + f[u] + f[v]; }print(ans), putc('\n');}return 0; }

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