题面

洛谷
bzoj

题解

离线处理+LCT

有点像星球大战
我们可以倒着做，断边变成连边

我们可以把边变成一个点

连边时，如果两个点本身不联通，就\(val\)赋为\(1\)，并连接这条边
如果，两个点本身就联通，那么就不连接这条边，把两点之间的\(val\)全部赋为\(0\)

\(ans\)就是求两点之间的\(sum(val)\)

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;

inline int gi() {
    RG int x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
    return f ? -x : x;
}

const int N = 52000;
struct node {
    int ch[2], f, sum, rev, v, ly;  
}t[N<<2];
bool isroot(int x) {
    return t[t[x].f].ch[0] != x && t[t[x].f].ch[1] != x;
}
inline int get(int x) {
    return t[t[x].f].ch[1] == x;
}
inline void pushup(int x) {
    t[x].sum = t[t[x].ch[0]].sum+t[t[x].ch[1]].sum+t[x].v;
}
void rotate(int x) {
    int y = t[x].f, z = t[y].f, k = get(x);
    if (!isroot(y))
        t[z].ch[get(y)] = x;
    t[x].f = z;
    t[t[x].ch[k^1]].f = y; t[y].ch[k] = t[x].ch[k^1];
    t[y].f = x; t[x].ch[k^1] = y;
    pushup(y);
    return ;
}
int top, S[N<<2];
inline void putly(int x) {t[x].ly = 1; t[x].sum = t[x].v = 0;}
void pushdown(int x) {
    if (t[x].rev) {
        swap(t[x].ch[0], t[x].ch[1]);
        if (t[x].ch[0]) t[t[x].ch[0]].rev ^= 1;
        if (t[x].ch[1]) t[t[x].ch[1]].rev ^= 1;
        t[x].rev = 0;
    }
    if (t[x].ly) {
        if (t[x].ch[0]) putly(t[x].ch[0]);
        if (t[x].ch[1]) putly(t[x].ch[1]);
        t[x].ly = 0;
    }
    return ;
}
void splay(int x) {
    S[top=1] = x;
    for (int i = x; !isroot(i); i = t[i].f) S[++top] = t[i].f;
    for (int i = top; i; i--) pushdown(S[i]);
    while (!isroot(x)) {
        int y = t[x].f;
        if (!isroot(y))(get(x)^get(y)) ? rotate(x):rotate(y);
        rotate(x);
    }
    pushup(x);
    return ;
}
void access(int x) {for (int y=0;x;y=x,x=t[x].f) splay(x),t[x].ch[1]=y,pushup(x);}
void makeroot(int x){access(x),splay(x),t[x].rev ^= 1;}
void split(int x, int y){makeroot(x),access(y),splay(y);}
void link(int x, int y) {makeroot(x);t[x].f = y;}
int findroot(int x) {access(x); splay(x);while (t[x].ch[0]) x = t[x].ch[0];return x;}

struct Line {
    int u, v;
    bool flag;
}E[N<<2];
map<pair<int, int>, int> M;
struct question {
    int op, ans, u, v;
}q[N];

int main() {
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    int n = gi(), m = gi();
    for (int i = 1; i <= m; i++) {
        int u = gi(), v = gi();
        if (u > v) swap(u, v);
        E[i] = (Line) {u, v, 0};
        t[i+n].v = 1;
        M[make_pair(u, v)] = i;
    }
    int cnt = 0;
    for (;;) {
        int c = gi(); if (c == -1) break;
        int u = gi(), v = gi(); if (u > v) swap(u, v);
        q[++cnt] = (question) {c, 0, u, v};
        if (!c) E[q[cnt].ans = M[make_pair(u, v)]].flag = 1;
    }
    for (int i = 1; i <= m; i++)
        if (!E[i].flag) {
            int u = E[i].u, v = E[i].v;
            if (findroot(u) != findroot(v))
                link(u, i+n), link(v, i+n);
            else {
                split(u, v);
                putly(v);
            }           
        }
    for (int i = cnt; i; i--) {
        if (q[i].op) {
            int u = q[i].u, v = q[i].v;
            split(u, v);            
            q[i].ans = t[v].sum;
        }
        else {
            int u = q[i].u, v = q[i].v;
            if (findroot(u) != findroot(v))
                link(u, q[i].ans+n), link(v, q[i].ans+n);
            else {
                split(u, v);
                putly(v);
            }
        }
    }
    for (int i = 1; i <= cnt; i++)
        if (q[i].op)
            printf("%d\n", q[i].ans);
    return 0;
}