原地址:http://www.cnblogs.com/Quinte/p/4898410.html
1.对节点X到Y的最短路上所有点权都加一个数W, 查询某个点的权值
解 :
这个操作等价于
a. 对X到根节点路径上所有点权加W
b. 对Y到根节点路径上所有点权加W
c. 对LCA(x, y)到根节点路径上所有点权值减W
d. 对LCA(x,y)的父节点 parent(LCA(x, y))到根节点路径上所有权值减W
于是要进行四次这样从一个点到根节点的区间修改
将问题进一步简化, 进行一个点X到根节点的区间修改, 查询其他一点Y时
只有X在Y的子树内, X对Y的值才有贡献且贡献值为W
于是只需要更新四个点, 查询一个点的子树内所有点权的和即可
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXN = 1e5 + 10;
vector<int> edge[MAXN];
int s[2 * MAXN];
int seq[2 * MAXN];
int seq1[2 * MAXN];
int depth[2 * MAXN];
int first[MAXN];
int dp[2 * MAXN][25];
int st[MAXN];
int ed[MAXN];
int parent[MAXN];
int cnt, num;
int Lowbit(int x) {
return x & (-x);
}
void Add(int x, int val, int n) {
if (x <= 0) return;
for (int i = x; i <= n; i += Lowbit(i)) {
s[i] += val;
}
}
int Sum(int x) {
int res = 0;
for (int i = x; i > 0; i -= Lowbit(i)) {
res += s[i];
}
return res;
}
void Dfs(int u, int fa, int dep) {
parent[u] = fa;
seq[++cnt] = u;
seq1[++num] = u;
first[u] = num;
depth[num] = dep;
st[u] = cnt;
int len = edge[u].size();
for (int i = 0; i < len; i++) {
int v = edge[u][i];
if (v != fa) {
Dfs(v, u, dep + 1);
seq1[++num] = u;
depth[num] = dep;
}
}
seq[++cnt] = u;
ed[u] = cnt;
}
void RMQ_Init(int n) {
for (int i = 1; i <= n; i++) {
dp[i][0] = i;
}
for (int j = 1; (1 << j) <= n; j++) {
for (int i = 1; i + (1 << j) - 1 <= n; i++) {
int a = dp[i][j - 1], b = dp[i + (1 << (j - 1))][j - 1];
dp[i][j] = depth[a] < depth[b] ? a : b;
}
}
}
int RMQ_Query(int l, int r) {
int k = 0;
while ((1 << (k + 1)) <= r - l + 1) k++;
int a = dp[l][k], b = dp[r - (1 << k) + 1][k];
return depth[a] < depth[b] ? a : b;
}
int LCA(int u, int v) {
int a = first[u], b = first[v];
if (a > b) a ^= b, b ^= a, a ^= b;
int res = RMQ_Query(a, b);
return seq1[res];
}
void Init(int n) {
for (int i = 0; i <= n; i++) {
edge[i].clear();
}
memset(s, 0, sizeof(s));
}
int main() {
int n, op;
int u, v, w;
int cmd;
while (scanf("%d %d", &n, &op) != EOF) {
Init(n);
for (int i = 0; i < n - 1; i++) {
scanf("%d %d", &u, &v);
edge[u].push_back(v);
edge[v].push_back(u);
}
cnt = 0, num = 0;
Dfs(1, -1, 0);
RMQ_Init(num);
while (op--) {
scanf("%d", &cmd);
if (cmd == 0) {
scanf("%d %d %d", &u, &v, &w);
int lca = LCA(u, v);
Add(st[u], w, cnt);
Add(st[v], w, cnt);
Add(lca, -w, cnt);
Add(parent[lca], -w, cnt);
} else if (cmd == 1) {
scanf("%d", &u);
printf("%d\n", Sum(ed[u]) - Sum(st[u] - 1));
}
}
}
return 0;
}