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SPOJ QTREE Query on a tree(树链剖分)

2014-09-30 22:02 369 查看


SPOJ QTREE Query on a tree

题目链接

树链剖分基础题

代码:

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 10005;

int dep
, top
, sz
, son
, fa
, id
, idx, val
;
int n;
vector<int> g
;

struct Edge {
int u, v, val;
void read() {
scanf("%d%d%d", &u, &v, &val);
}
} e
;

void dfs1(int u, int f, int d) {
dep[u] = d;
sz[u] = 1;
son[u] = 0;
fa[u] = f;
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == f) continue;
dfs1(v, u, d + 1);
sz[u] += sz[v];
if (sz[son[u]] < sz[v])
son[u] = v;
}
}

void dfs2(int u, int tp) {
top[u] = tp;
id[u] = ++idx;
if (son[u]) dfs2(son[u], tp);
for (int i = 0; i < g[u].size(); i++) {
int v = g[u][i];
if (v == fa[u] || v == son[u]) continue;
dfs2(v, v);
}
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)
const int INF = 0x3f3f3f3f;

struct Node {
int l, r, val;
} node[N * 4];

void pushup(int x) {
node[x].val = max(node[lson(x)].val, node[rson(x)].val);
}

void build(int l, int r, int x = 0) {
node[x].l = l; node[x].r = r;
if (l == r) {
node[x].val = val[l];
return;
}
int mid = (l + r) / 2;
build(l, mid, lson(x));
build(mid + 1, r, rson(x));
pushup(x);
}

void add(int v, int val, int x = 0) {
if (node[x].l == node[x].r) {
node[x].val = val;
return;
}
int mid = (node[x].l + node[x].r) / 2;
if (v <= mid) add(v, val, lson(x));
if (v > mid) add(v, val, rson(x));
pushup(x);
}

int query(int l, int r, int x = 0) {
if (node[x].l >= l && node[x].r <= r) {
return node[x].val;
}
int mid = (node[x].l + node[x].r) / 2;
int ans = 0;
if (l <= mid) ans = max(ans, query(l, r, lson(x)));
if (r > mid) ans = max(ans, query(l, r, rson(x)));
return ans;
}

int gao(int u, int v) {
int tp1 = top[u], tp2 = top[v];
int ans = 0;
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
ans = max(query(id[tp1], id[u]), ans);
u = fa[tp1];
tp1 = top[u];
}
if (u == v) return ans;
if (dep[u] > dep[v]) swap(u, v);
ans = max(query(id[son[u]], id[v]), ans);
return ans;
}

int T;

int main() {
scanf("%d", &T);
while (T--) {
scanf("%d", &n);
idx = 0;
for (int i = 1; i <= n; i++) g[i].clear();
for (int i = 1; i < n; i++) {
e[i].read();
g[e[i].u].push_back(e[i].v);
g[e[i].v].push_back(e[i].u);
}
dfs1(1, 0, 1);
dfs2(1, 1);
char Q[10];
for (int i = 1; i < n; i++) {
if (dep[e[i].u] < dep[e[i].v]) swap(e[i].u, e[i].v);
val[id[e[i].u]] = e[i].val;
}
build(1, idx);
int a, b;
while (scanf("%s", Q)) {
if (Q[0] == 'D') break;
scanf("%d%d", &a, &b);
if (Q[0] == 'Q') printf("%d\n", gao(a, b));
if (Q[0] == 'C') add(id[e[a].u], b);
}
}
return 0;
}
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