[BZOJ2631][Tree][LCT]

题目大意：

给定一棵N<=100000个节点的无根树，有四种操作：删除一条边后再加一条边（保证形成的还是一棵树）、在树上的一条路径上全部加一个权值、在树上的一条路径上全部乘一个权值、求树上一条路径的权值和。

答案要 %51061

思路：

显然这棵树是动态的，那么用LCT来维护连通性就好了，加和乘的操作可以在splay里面打lazy标记来维护。

求树上的一条路径可以用split操作。

注意在access的时候对于每个经过的点都要pushup一遍。

代码：

#include <bits/stdc++.h>using namespace std;typedef unsigned int ll;const int Maxn = 100005;const int Mod = 51061;namespace IO {    inline char get(void) {        static char buf[1000000], *p1 = buf, *p2 = buf;        if (p1 == p2) {            p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);            if (p1 == p2) return EOF;        }        return *p1++;    }    inline void read(int &x) {        x = 0; static char c; bool f = 0;        for (; !(c >= '0' && c <= '9'); c = get()) if (c == '-') f = 1;        for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = get()); if (f) x = -x;    }    inline void read(char &x) {        x = get();        while (!(x == '*' || x == '/' || x == '+' || x == '-')) x = get();    }    inline void write(int x) {        if (!x) return (void)puts("0");        if (x < 0) putchar('-'), x = -x;        static short s[12], t;        while (x) s[++t] = x % 10, x /= 10;        while (t) putchar('0' + s[t--]);        putchar('\n');    }};int fa[Maxn], c[Maxn][2], st[Maxn], top, size[Maxn], nxt[Maxn];bool rev[Maxn];ll mx[Maxn], at[Maxn], val[Maxn], sum[Maxn];inline void cal(int x, int m, int a) {    if (!x) return;    val[x] = (val[x] * m + a) % Mod;    sum[x] = (sum[x] * m + a * size[x]) % Mod;    at[x] = (at[x] * m + a) % Mod;    mx[x] = (mx[x] * m) % Mod;}inline bool isRt(int x) {    return c[fa[x]][0] != x && c[fa[x]][1] != x;}inline void pushDown(int x) {    static int l, r;    l = c[x][0], r = c[x][1];    if (rev[x]) {        rev[l] ^= 1; rev[r] ^= 1;        swap(c[x][0], c[x][1]);        rev[x] ^= 1;    }    int m = mx[x], a = at[x];    mx[x] = 1, at[x] = 0;    if (m != 1 || a != 0) {        cal(l, m, a), cal(r, m, a);    }}inline void pushUp(int x) {    int l = c[x][0], r = c[x][1];    sum[x] = (sum[l] + sum[r] + val[x]) % Mod;    size[x] = (size[l] + size[r] + 1) % Mod;;}inline void rotate(int x) {    int y = fa[x], z = fa[y], l, r;    r = c[y][0] == x; l = r ^ 1;    if (!isRt(y)) {        if (c[z][0] == y) c[z][0] = x;        else c[z][1] = x;    }    fa[x] = z; fa[y] = x; fa[c[x][r]] = y;    c[y][l] = c[x][r]; c[x][r] = y;    pushUp(y); pushUp(x);}inline void splay(int x) {    st[top = 1] = x;    for (int i = x; !isRt(i); i = fa[i]) st[++top] = fa[i];    while (top) pushDown(st[top--]);    int y, z;    while (!isRt(x)) {        y = fa[x], z = fa[y];        if (!isRt(y)) {            if (c[y][0] == x ^ c[z][0] == y) rotate(x);            else rotate(y);        }        rotate(x);    }} inline void access(int x) {    int t = 0;    while (x) {        splay(x);        c[x][1] = t;        pushUp(x);        t = x; x = fa[x];    }}inline void rever(int x) {    access(x); splay(x); rev[x] ^= 1;}inline void link(int x, int y) {    rever(x); fa[x] = y;    // splay(x);}inline void cut(int x, int y) {    rever(x); access(y); splay(y); c[y][0] = fa[x] = 0;}inline void split(int x, int y) {    rever(y);     access(x);     splay(x);}int n, m;int main(void) {    //freopen("in.txt", "r", stdin);    IO::read(n), IO::read(m);    for (int i = 1; i <= n; i++) val[i] = sum[i] = mx[i] = size[i] = 1;    for (int i = 1, u, v; i < n; i++) {        IO::read(u), IO::read(v);        link(u, v);    }    char op; int u, v, c, d;    for (int i = 1; i <= m; i++) {        IO::read(op); IO::read(u), IO::read(v);        if (op == '+') {            IO::read(c);            split(u, v); cal(u, 1, c);        } else if (op == '-') {            IO::read(c), IO::read(d);            cut(u, v), link(c, d);        } else if (op == '*') {            IO::read(c);            split(u, v), cal(u, c, 0);        } else {            split(u, v); IO::write(sum[u]);        }    }    return 0;}

完。

By g1n0st