LCA + 树状数组 + 树上RMQ

题目链接：http://poj.org/problem?id=2763

思路：首先求出树上dfs序列，并且标记树上每个节点开始遍历以及最后回溯遍历到的时间戳，由于需要修改树上的某两个节点之间的权值，如果parent[v] = u, 那么说明修改之后的v的子树到当前根的距离都会改变，由于遍历到v时有开始时间戳以及结束时间戳，那么处于这个区间所有节点都会影响到，于是我们可以通过数组数组来更新某个区间的值，只需令从区间起始点之后的每个值都增加一个改变量了，令区间中止点之后的每个值都减小一个改变量，这样我们就可以得到处于该区间的值的改变量。至于如何求树上两个节点的LCA,这里可以使用RMQ,O(n * log(n))的预处理，O(1)的查询复杂度。\

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX_N = (200000 + 10000);struct Edge {    int u, v, w, next;    Edge () {}    Edge (int _u, int _v, int _w, int _next) : u(_u), v(_v), w(_w), next(_next) {}} EE[MAX_N << 2], edge[MAX_N << 2];int NE, head[MAX_N];void Init(){    NE = 0;    memset(head, -1, sizeof(head));}void Insert(int u, int v, int w){    edge[NE].u = u;    edge[NE].v = v;    edge[NE].w = w;    edge[NE].next = head[u];    head[u] = NE++;}int index, dfs_seq[MAX_N << 2]; //dfs序列int First[MAX_N], End[MAX_N]; //First数组表示第一次遍历到的时间戳，End数组表示最后回溯时遍历到的时间戳int parent[MAX_N], dep[MAX_N << 2]; //深度int N, Q, st, cost[MAX_N];void dfs(int u, int fa, int d, int c){    parent[u] = fa;    First[u] = index;    dfs_seq[index] = u;    dep[index++] = d;    cost[u] = c;    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].v, w = edge[i].w;        if (v == fa) continue;        dfs(v, u, d + 1, cost[u] + w);        dfs_seq[index] = u;        dep[index++] = d;    }    End[u] = index;}int dp[MAX_N][40]; //dp[i][j]表示从时间戳i开始，长度为(1 << j)的区间中深度最小的点的时间戳void Init_RMQ(){    for (int i = 0; i < index; ++i) dp[i][0] = i;    for (int j = 1; (1 << j) < index; ++j) {        for (int i = 0; i + (1 << j) - 1 < index; ++i) {            dp[i][j] = (dep[ dp[i][j - 1] ] < dep[ dp[i + (1 << (j - 1))][j - 1] ] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1]);        }    } } int RMQ_Query(int l, int r) {     int k = (int)(log(r * 1.0 - l + 1) / log(2.0));     return (dep[ dp[l][k] ] < dep[ dp[r - (1 << k) + 1][k] ] ? dp[l][k] : dp[r - (1 << k) + 1][k]); } int LCA(int x, int y) {     if (x > y) swap(x, y);     return dfs_seq[RMQ_Query(x, y)]; }int C[MAX_N << 2];int lowbit(int x){    return x & (-x);}void update(int i, int val){    while (i < index) {        C[i] += val;        i += lowbit(i);    }}int getSum(int i){    int sum = 0;    while (i > 0) {        sum += C[i];        i -= lowbit(i);    }    return sum;} int main() {     while (~scanf("%d %d %d", &N, &Q, &st)) {        Init();        for (int i = 1; i < N; ++i) {            int u, v, w;            scanf("%d %d %d", &u, &v, &w);            Insert(u, v, w);            Insert(v, u, w);            EE[i] = Edge(u, v, w, -1);        }        index = 0;        dfs(st, st, 0, 0);        Init_RMQ();        memset(C, 0, sizeof(C));        for (int i = 1; i <= Q; ++i) {            int opt;            scanf("%d", &opt);            if (opt == 0) {                int ed, fa, x, y, z;                scanf("%d", &ed);                x = First[st], y = First[ed], z = First[fa = LCA(x, y)];                printf("%d\n", getSum(x + 1) + getSum(y + 1) - 2 * getSum(z + 1) + cost[st] + cost[ed] - 2 * cost[fa]);                st = ed;            } else {                int id, w;                scanf("%d %d", &id, &w);                int u = EE[id].u, v = EE[id].v, tmp = w - EE[id].w;                EE[id].w = w;                if (parent[u] == v) swap(u, v);                update(First[v] + 1, tmp);                update(End[v] + 1, -tmp);            }        }     }     return 0; }