loj10139. 「一本通 4.5 练习 1」树上操作（loj2125. 「HAOI2015」树上操作 ）

思路：

　　第二遍dfs时记录end[x]为在结点序列中以x为根的子树最后访问的节点，写个线段树标记下传即可。与值有关的数据注意long long

#include<cstdio>
#include<iostream>
#include<vector>
#include<cmath>
#include<string>
using namespace std;
const int maxn = 100010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar(), flag = 0;
    while(ch < ‘0‘ || ch > ‘9‘)    {
        if(ch == ‘-‘)    flag = 1;
        ch = getchar();
    }
    while(ch >= ‘0‘ && ch <= ‘9‘)
        x = 10 * x + ch - 48, ch = getchar();
    if(flag)    x = -x;
}
int n, m, rt = 1;
int val[maxn];
int head[maxn];
int go[maxn << 1];
int nxt[maxn << 1];

int f[maxn];
int deep[maxn];
int size[maxn];
int son[maxn];
int top[maxn];
int seg[maxn];
int rev[maxn];
int end[maxn];
long long cov[maxn << 2];
long long sum[maxn << 2];
long long SUM;
inline void init(){
    qread(n);        qread(m);
    for(int i=1; i<=n; ++i)
        qread(val[i]);
    for(register int i=1; i<n; ++i){
        register int x, y;
        qread(x), qread(y);
        go[i] = y;
        go[i + n] = x;
        nxt[i] = head[x];
        head[x] = i;
        nxt[i + n] = head[y];
        head[y] = i + n;
    }
    deep[rt] = 1;
    seg[0] = seg[1] = rev[1] = top[1] = 1;
}
void dfs1(int x){
    size[x] = 1;
    for(register int i = head[x]; i; i = nxt[i])
        if(!deep[go[i]]){
            f[go[i]] = x;
            deep[go[i]] = deep[x] + 1;
            dfs1(go[i]);
            size[x] += size[go[i]];
            if(size[go[i]] > size[son[x]])
                son[x] = go[i];
        }
}
void dfs2(int x){
    end[x] = x;
    if(son[x]){
        seg[son[x]] = ++seg[0];
        top[son[x]] = top[x];
        rev[seg[0]] = son[x];
        dfs2(son[x]);
        end[x] = end[son[x]];
    }
    for(register int i = head[x]; i; i = nxt[i]){
        if(!top[go[i]]){
            seg[go[i]] = ++seg[0];
            rev[seg[0]] = go[i];
            top[go[i]] = go[i];
            dfs2(go[i]);
            end[x] = end[go[i]];
        }
    }
}
inline void pushdown(int x, int t){
    cov[x << 1] += cov[x];
    cov[x << 1 | 1] += cov[x];
    sum[x << 1] += cov[x] * ((t + 1) >> 1);
    sum[x << 1 | 1] += cov[x] * (t >> 1);
    cov[x] = 0;
}
void segbuild(int l, int r, int x){
    if(l==r){
        sum[x] = val[rev[l]];
        return ;
    }
    int mid = (l + r) >> 1;
    segbuild(l, mid, x << 1);
    segbuild(mid + 1, r, x << 1 | 1);
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void segadd(int l, int r, int L, int R, long long v, int x){
    if(L <= l && R >= r){
        cov[x] += v;
        sum[x] += v * (r - l + 1);
        return ;
    }
    pushdown(x, r - l + 1);
    int mid = (l + r) >> 1;
    if(mid >= L)    segadd(l, mid, L, R, v, x << 1);
    if(mid < R)        segadd(mid + 1, r, L, R, v, x << 1 | 1);
    sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void segquery(int l, int r, int L, int R, int x){
    if(L <= l && R >= r){
        SUM += sum[x];
        return ;
    }
    pushdown(x, r - l + 1);
    int mid = (l + r) >> 1;
    if(mid >= L)    segquery(l, mid, L, R, x << 1);
    if(mid < R)        segquery(mid + 1, r, L, R,  x << 1 | 1);
}
void treeask(int x, int y){
    int fx = top[x], fy = top[y];
    while(fx != fy){
        if(deep[fx] < deep[fy])    swap(x, y), swap(fx, fy);
        segquery(1, seg[0], seg[fx], seg[x], 1);
        x = f[fx];
        fx = top[x];
    }
    if(deep[x] > deep[y])    swap(x, y);
    segquery(1, seg[0], seg[x], seg[y], 1);
}
int main(void){
    freopen("data6.in.txt", "r", stdin);
    freopen("data6my.txt", "w", stdout);
    init();
    dfs1(rt);
    dfs2(rt);
    segbuild(1, seg[0], 1);
    while(m--){
        int op;
        qread(op);
        if(op == 1){
            int x, a;
            qread(x), qread(a);
            segadd(1, seg[0], seg[x], seg[x], a, 1);
        }
        else if(op == 2){
            int x, a;
            qread(x), qread(a);
            segadd(1, seg[0], seg[x], seg[end[x]], a, 1);
        }
        else{
            int x;
            qread(x);
            SUM = 0;
            treeask(rt, x);
            printf("%lld\n", SUM);
        }
    }
}

原文地址：https://www.cnblogs.com/junk-yao-blog/p/9480761.html