您的位置:首页 > 其它

poj 2763(树链剖分)

2015-10-15 22:20 357 查看
题意:有一个人在一棵树的某个节点s,然后给出了树的每条边都有一个权值,有两种操作,0 a 是问从节点s到a的路径权值和,然后a就成了s,1 a b把第a条边权值变为b。

题解:边权转点权的模板题。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100005;
struct Edge {
int u, v, w, nxt;
Edge() {}
Edge(int a, int b, int c, int d): u(a), v(b), w(c), nxt(d) {}
}e[N << 1];
int size
, top
, son
, fa
, dep
, id
;
int n, q, s, head
, cnt, tot, tree[N << 2];

void AddEdge(int u, int v, int w) {
e[cnt] = Edge(u, v, w, head[u]);
head[u] = cnt++;
e[cnt] = Edge(v, u, w, head[v]);
head[v] = cnt++;
}

void dfs2(int u, int tp) {
top[u] = tp;
id[u] = ++tot;
if (son[u]) dfs2(son[u], tp);
for (int i = head[u]; i + 1; i = e[i].nxt) {
int v = e[i].v;
if (v == fa[u] || v == son[u]) continue;
dfs2(v, v);
}
}

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

void pushup(int k) {
tree[k] = tree[k * 2] + tree[k * 2 + 1];
}

void modify(int k, int left, int right, int pos, int v) {
if (left == right) {
tree[k] = v;
return;
}
int mid = (left + right) / 2;
if (pos <= mid)
modify(k * 2, left, mid, pos, v);
else
modify(k * 2 + 1, mid + 1, right, pos, v);
pushup(k);
}

int query(int k, int left, int right, int l, int r) {
if (l <= left && right <= r)
return tree[k];
int mid = (left + right) / 2, res = 0;
if (l <= mid)
res += query(k * 2, left, mid, l, r);
if (r > mid)
res += query(k * 2 + 1, mid + 1, right, l, r);
return res;
}

void solve(int u, int v) {
int tp1 = top[u], tp2 = top[v];
int res = 0;
while (tp1 != tp2) {
if (dep[tp1] < dep[tp2]) {
swap(tp1, tp2);
swap(u, v);
}
res += query(1, 1 ,tot, id[tp1], id[u]);
u = fa[tp1];
tp1 = top[u];
}
if (u != v) {
if (dep[u] > dep[v]) swap(u, v);
res += query(1, 1, tot, id[son[u]], id[v]);
}
printf("%d\n", res);
}

int main() {
while (scanf("%d%d%d", &n, &q, &s) == 3) {
int u, v, w;
memset(head, -1, sizeof(head));
memset(tree, 0, sizeof(tree));
cnt = tot = 0;
for (int i = 0; i < n - 1; i++) {
scanf("%d%d%d", &u, &v, &w);
AddEdge(u, v, w);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for (int i = 0; i < cnt; i += 2) {
u = e[i].u, v = e[i].v;
if (dep[u] < dep[v]) swap(u, v);
modify(1, 1, tot, id[u], e[i].w);
}
int op, a, b;
while (q--) {
scanf("%d%d", &op, &a);
if (op == 0) {
solve(s, a);
s = a;
}
else {
scanf("%d", &b);
u = e[a - 1 << 1].u;
v = e[a - 1 << 1].v;
if (dep[u] < dep[v]) swap(u, v);
modify(1, 1, tot, id[u], b);
}
}
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: