HDU 6162 Ch’s gift(树链剖分+线段树)

题意：给你一颗树，每个节点有一个值，q次询问，每次询问u到v的路径上节点权值在a至b的节点的权值和。

n，q<=1e5

思路：可以离线处理，分别求出路径中<a的和，<=b的和，两者做差就是答案。

代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 1e5+5;int deep[maxn], fa[maxn], top[maxn], tree[maxn], pre[maxn], son[maxn], num[maxn];int n, q, tot;ll treeSum[maxn*4], ansL[maxn], ansR[maxn];vector<int> g[maxn];struct node1{    int x, id;    bool operator < (const node1 &a) const    {        return x < a.x;    }}a[maxn];struct node2{    int x, y, a, b, id;}op[maxn];bool cmp1(const node2 &a, const node2 &b){    return a.a < b.a;}bool cmp2(const node2 &a, const node2 &b){    return a.b < b.b;}void init(){    tot = 0;    memset(son, 0, sizeof(son));    memset(num, 0, sizeof(num));    for(int i = 0; i <= n; i++)        g[i].clear();}void dfs1(int u, int pre, int d){    deep[u] = d;    fa[u] = pre;    num[u] = 1;    for(int i = 0, len = g[u].size(); i < len; i++)    {        int v = g[u][i];        if(v == pre) continue;        dfs1(v, u, d+1);        num[u] += num[v];        if(!son[v] || num[v] > num[son[u]])            son[u] = v;    }}void dfs2(int u, int tp){    top[u] = tp;    tree[u] = ++tot;    pre[tree[u]] = u;    if(!son[u]) return ;    dfs2(son[u], tp);    for(int i = 0, len = g[u].size(); i < len; i++)    {        int v = g[u][i];        if(v != son[u] && v != fa[u])            dfs2(v, v);    }}void push_up(int root){    treeSum[root] = treeSum[root*2]+treeSum[root*2+1];}void update(int root, int l, int r, int pos, int val){    if(l == r)    {        treeSum[root] += val;        return ;    }    int mid = (l+r)/2;    if(pos <= mid) update(root*2, l, mid, pos, val);    else update(root*2+1, mid+1, r, pos, val);    push_up(root);}ll query(int root, int l, int r, int i, int j){    if(i <= l && j >= r) return treeSum[root];    int mid = (l+r)/2;    ll res = 0;    if(i <= mid) res += query(root*2, l, mid, i, j);    if(j > mid) res += query(root*2+1, mid+1, r, i, j);    return res;}ll ask(int x, int y){    int f1 = top[x], f2 = top[y];    ll ans = 0;    while(f1 != f2)    {        if(deep[f1] < deep[f2]) swap(f1, f2), swap(x, y);        ans += query(1, 1, n, tree[f1], tree[x]);        x = fa[f1], f1 = top[x];    }    ans += (deep[x]>deep[y]) ? query(1, 1, n, tree[y], tree[x]) : query(1, 1, n, tree[x], tree[y]);    return ans;}int main(void){    while(cin >> n >> q)    {        init();        for(int i = 1; i <= n; i++)            scanf("%d", &a[i].x), a[i].id = i;        for(int i = 1; i < n; i++)        {            int u, v;            scanf("%d%d", &u, &v);            g[u].push_back(v);            g[v].push_back(u);        }        dfs1(1, 0, 1); dfs2(1, 1);        for(int i = 1; i <= q; i++)            scanf("%d%d%d%d", &op[i].x, &op[i].y, &op[i].a, &op[i].b), op[i].id = i;        sort(a+1, a+1+n);        //L        memset(treeSum, 0, sizeof(treeSum));        sort(op+1, op+1+q, cmp1);        for(int i = 1, j = 1; i <= q; i++)        {            while(j <= n && a[j].x < op[i].a)            {                update(1, 1, n, tree[a[j].id], a[j].x);                j++;            }            ansL[op[i].id] = ask(op[i].x, op[i].y);        }        //R        memset(treeSum, 0, sizeof(treeSum));        sort(op+1, op+1+q, cmp2);        for(int i = 1, j = 1; i <= q; i++)        {            while(j <= n && a[j].x <= op[i].b)            {                update(1, 1, n, tree[a[j].id], a[j].x);                j++;            }            ansR[op[i].id] = ask(op[i].x, op[i].y);        }        for(int i = 1; i <= q; i++)        {            if(i != 1) printf(" ");            printf("%lld", ansR[i]-ansL[i]);        }        puts("");    }    return 0;}