2018焦作网络赛 E Jiu Yuan Wants to Eat（线段树+树链剖分）

题意

对一个有1e5个点，点权初值为0的树上进行4种操作：

1、结点u到结点v上的所有点权乘x。

2、结点u到结点v上所有的点权加x。

3、结点u到结点v上所有的点权取非。

4、结点u到结点v路径上点权的和。

答案模\(2^{64}\)

思路

对操作1、2树链剖分加线段树维护即可，对操作3，取非操作为：一个二进制位上都为1

的数减去当前值，即：\(2^{64}-1-x\) ， 由于答案模\(2^{64}\)，通过点权用unsigned long long，只需维护\(-(x+1)\) 操作即可。

Code

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e5+10;
const int SZ = 1 << 20;  //快速io
struct fastio {
    char inbuf[SZ];
    char outbuf[SZ];
    fastio() {
        setvbuf(stdin, inbuf, _IOFBF, SZ);
        setvbuf(stdout, outbuf, _IOFBF, SZ);
    }
} io;

void read(int &x) {
    x=0; char ch, c=getchar();
    while(c<'0' || c>'9') ch=c, c=getchar();
    while(c>='0'&&c<='9') x=x*10+c-'0', c=getchar();
    if(ch=='-') x=-x;
}

int n, m;
int dfn[maxn], rnk[maxn], top[maxn], tot;
int sz[maxn], son[maxn], dep[maxn], fa[maxn];
vector<int> g[maxn];

void dfs1(int x, int f) {
    sz[x]=1, son[x]=0, dep[x] = dep[f]+1;
    for (auto v: g[x]) {
        fa[v] = x;
        dfs1(v, x);
        sz[x] += sz[v];
        if(sz[v] > sz[son[x]]) son[x] = v;
    }
}
void dfs2(int x, int topf) {
    dfn[x]=++tot, rnk[tot]=x, top[x]=topf;
    if(!son[x]) return;
    dfs2(son[x], topf);
    for (auto v: g[x]) {
        if(v!=son[x])
            dfs2(v, v);
    }
}

typedef unsigned long long ull;
struct segTree {
    struct node {
        int l, r;
        ull val, add, mul;
        inline int length() { return r-l+1; }
    }tr[maxn<<2];
    void pushup(int i) {
        tr[i].val = tr[i<<1].val + tr[i<<1|1].val;
    }
    void pushdown(int i) {
        if(tr[i].add==0 && tr[i].mul==1) return;
        tr[i<<1].val = tr[i<<1].val*tr[i].mul + tr[i].add*(tr[i<<1].r-tr[i<<1].l+1);
        tr[i<<1|1].val = tr[i<<1|1].val*tr[i].mul + tr[i].add*(tr[i<<1|1].r-tr[i<<1|1].l+1);
        tr[i<<1].add = tr[i<<1].add*tr[i].mul+tr[i].add;
        tr[i<<1|1].add = tr[i<<1|1].add*tr[i].mul+tr[i].add;
        tr[i<<1].mul *= tr[i].mul;
        tr[i<<1|1].mul *= tr[i].mul;
        tr[i].add = 0, tr[i].mul = 1;
    }
    void build(int l, int r, int i) {
        tr[i] = node{l, r, 0, 0, 1};
        if(l==r) return;
        int mid = l+r>>1;
        build(l, mid, i<<1);
        build(mid+1, r, i<<1|1);
        pushup(i);
    }
    void update(int l, int r, ull val, int op, int i) {
        if(l<=tr[i].l && tr[i].r<=r) {
            if(op==2) {
                tr[i].val += val*tr[i].length();
                tr[i].add += val;
            } else {
                tr[i].val *= val;
                tr[i].mul *= val;
                tr[i].add *= val;
            }
            return;
        }
        int mid = tr[i].l+tr[i].r>>1;
        pushdown(i);
        if(l<=mid) update(l, r, val, op, i<<1);
        if(r>mid) update(l, r, val, op, i<<1|1);
        pushup(i);
    }
    ull query(int l, int r, int i) {
        if(l<=tr[i].l && tr[i].r<=r) return tr[i].val;
        int mid=tr[i].l+tr[i].r>>1;
        ull res=0;
        pushdown(i);
        if(l<=mid) res += query(l, r, i<<1);
        if(r>mid) res += query(l, r, i<<1|1);
        return res;
    }
}st;

void update(int u, int v, ull val, int op) {
    while(top[u] != top[v]) {
        if(dep[top[u]] < dep[top[v]]) swap(u, v);
        st.update(dfn[top[u]], dfn[u], val, op, 1);
        u = fa[top[u]];
    }
    if(dep[u] > dep[v]) swap(u, v);
    st.update(dfn[u], dfn[v], val, op, 1);
}
ull query(int u, int v) {
    ull res = 0;
    while(top[u] != top[v]) {
        if(dep[top[u]] < dep[top[v]]) swap(u, v);
        res += st.query(dfn[top[u]], dfn[u], 1);
        u = fa[top[u]];
    }
    if(dep[u] > dep[v]) swap(u, v);
    res += st.query(dfn[u], dfn[v], 1);
    return res;
}

int main() {
//    freopen("input.in", "r", stdin);
    while(~scanf("%d", &n)) {
        tot = 0;
        for (int i=1; i<=n; ++i) g[i].clear();
        for (int x, i=2; i<=n; ++i) {
            read(x);
            g[x].push_back(i);
        }
        st.build(1, n, 1);
        dfs1(1, 0);
        dfs2(1, 1);
//        for (int i=1; i<=n; ++i) printf("%d ", top[i]); puts("");
//        for (int i=1; i<=n; ++i) printf("%d ", dep[i]); puts("");
        scanf("%d", &m);
        for (int op, u, v, i=1; i<=m; ++i) {
            read(op); read(u); read(v);
//            scanf("%d%d%d", &op, &u, &v);
            if(op==1 || op==2) {
                ull x;
                scanf("%llu", &x);
                update(u, v, x, op);
            } else if(op==3) {
                update(u, v, 1, 2);
                update(u, v, -1, 1);
            } else {
                ull ans = query(u, v);
                printf("%llu\n", ans);
            }
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/acerkoo/p/11295766.html