spoj 375. Query on a tree(树链剖分)

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题目链接：spoj 375. Query on a tree

题目大意：

poj 3237的简化版，用同一份代码都能过。

解题思路：略。

#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;
const int maxn = 10005;
const int INF = 0x3f3f3f3f;#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int val[maxn];
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];
inline void pushup(int u) {s[u] = max(s[lson(u)], s[rson(u)]);
}void build (int u, int l, int r) {lc[u] = l;rc[u] = r;if (l == r) {s[u] = val[l];return;}int mid = (l + r) / 2;build(lson(u), l, mid);build(rson(u), mid + 1, r);pushup(u);
}void modify(int u, int x, int w) {if (lc[u] == x && rc[u] == x) {s[u] = w;return;}int mid = (lc[u] + rc[u]) / 2;if (x <= mid)modify(lson(u), x, w);elsemodify(rson(u), x, w);pushup(u);
}int query (int u, int l, int r) {if (l <= lc[u] && rc[u] <= r)return s[u];int mid = (lc[u] + rc[u]) / 2, ret = -INF;if (l <= mid)ret = max(ret, query(lson(u), l, r));if (r > mid)ret = max(ret, query(rson(u), l, r));pushup(u);return ret;
}int N, ne, first[maxn], jump[maxn * 2];
int id, idx[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn], dep[maxn];
struct Edge {int u, v, w;void set(int u, int v, int w) {this->u = u;this->v = v;this->w = w;}
}ed[maxn * 2];inline void add_Edge(int u, int v, int w) {ed[ne].set(u, v, w);jump[ne] = first[u];first[u] = ne++;
}void dfs (int u, int pre, int d) {far[u] = pre;dep[u] = d;son[u] = 0;cnt[u] = 1;for (int i = first[u]; i + 1; i = jump[i]) {int v = ed[i].v;if (v == pre)continue;dfs(v, u, d + 1);cnt[u] += cnt[v];if (cnt[son[u]] < cnt[v])son[u] = v;}
}void dfs (int u, int rot) {top[u] = rot;idx[u] = ++id;if (son[u])dfs(son[u], rot);for (int i = first[u]; i + 1; i = jump[i]) {int v = ed[i].v;if (v == far[u] || v == son[u])continue;dfs(v, v);}
}void init () {int u, v, w;scanf("%d", &N);ne = id = 0;memset(first, -1, sizeof(first));for (int i = 0; i < N - 1; i++) {scanf("%d%d%d", &u, &v, &w);add_Edge(u, v, w);add_Edge(v, u, w);}dfs(1, 0, 0);dfs(1, 1);for (int i = 0; i < N - 1; i++) {int k = i * 2;if (dep[ed[k].u] < dep[ed[k].v])swap(ed[k].u, ed[k].v);val[idx[ed[k].u]] = ed[k].w;}build(1, 1, N);
}int solve (int u, int v) {int p = top[u], q = top[v], ret = -INF;while (p != q) {if (dep[p] < dep[q]) {swap(p, q);swap(u, v);}ret = max(ret, query(1, idx[p], idx[u]));u = far[p];p = top[u];}if (u == v)return ret;if (dep[u] > dep[v])swap(u, v);ret = max(ret, query(1, idx[son[u]], idx[v]));return ret;
}int main () {int cas;scanf("%d", &cas);while (cas--) {init();int u, v;char op[20];while (scanf("%s", op), strcmp(op, "DONE") != 0) {scanf("%d%d", &u, &v);if (op[0] == 'Q')printf("%d\n", solve(u, v));elsemodify(1, idx[ed[u*2-2].u], v);}}return 0;
}