洛谷 - P2542 [AHOI2005] 航线规划

传送门：洛谷 - P2542

思路分析

询问两点之间的关键边数量，在环里面的边肯定不是关键边，把关键边权值设为1
就是求两点之间的边权和，正向删边不好操作，可以逆向加边，对于一棵生成树，如果加边形成了环，那么把环上的边全部置为0就行了

AC代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a, b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror = 0;

    inline char nc() {
        static char buf[BUF_SIZE], *p1 = buf + BUF_SIZE, *pend = buf + BUF_SIZE;
        if (p1 == pend) {
            p1 = buf;
            pend = buf + fread(buf, 1, BUF_SIZE, stdin);
            if (pend == p1) {
                IOerror = 1;
                return -1;
            }
        }
        return *p1++;
    }

    inline bool blank(char ch) {
        return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
    }

    template<class T>
    inline bool read(T &x) {
        bool sign = 0;
        char ch = nc();
        x = 0;
        for (; blank(ch); ch = nc());
        if (IOerror)return false;
        if (ch == '-')sign = 1, ch = nc();
        for (; ch >= '0' && ch <= '9'; ch = nc())x = x * 10 + ch - '0';
        if (sign)x = -x;
        return true;
    }

    template<class T, class... U>
    bool read(T &h, U &... t) {
        return read(h) && read(t...);
    }

#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;

namespace LCT {
#define fa(x) (tree[x].fa)
#define ls(x) (tree[x].ch[0])
#define rs(x) (tree[x].ch[1])
#define ident(x,f) (rs(f)==x)
#define connect(x,f,s) tree[f].ch[s]=x,tree[x].fa=f
#define reverse(x) swap(ls(x),rs(x)),tree[x].tag^=1
#define notroot(x) (ls(fa(x)) == x || rs(fa(x)) == x)

    struct node {
        int fa,ch[2],val,res,lazy;
        bool tag;
    } tree[N];

    inline void pp(int x) {
        tree[x].res = tree[ls(x)].res + tree[rs(x)].res +tree[x].val;
    }

    inline void cover(int x){
        tree[x].val = 0;
        tree[x].res = 0;
        tree[x].lazy = 1;
    }

    inline void pushdown(int x) {
        if(tree[x].lazy){
            if(ls(x)) cover(ls(x));
            if(rs(x)) cover(rs(x));
        }
        if(tree[x].tag) {
            if(ls(x)) reverse(ls(x));
            if(rs(x)) reverse(rs(x));
        }
        tree[x].tag=0;
        tree[x].lazy=0;
    }

    inline void pushall(int x) {
        if(notroot(x)) pushall(fa(x));
        pushdown(x);
    }

    inline void rotate(int x) {
        int f = fa(x),ff=fa(f),fs=ident(x,f),ffs=ident(f,ff);
        connect(tree[x].ch[fs^1],f,fs);
        fa(x) = ff;
        if(notroot(f)) tree[ff].ch[ffs] = x;
        connect(f,x,fs^1);
        pp(f),pp(x);
    }

    inline void splaying(int x) {
        pushall(x);
        while(notroot(x)) {
            int f=fa(x),ff=fa(f);
            if(notroot(f)) ident(f,ff)^ident(x,f) ? rotate(x):rotate(f);
            rotate(x);
        }
    }

    inline void access(int x) {
        for(int y=0; x; x=fa(x)) {
            splaying(x);
            rs(x)=y;
            pp(x);
            y=x;
        }
    }

    inline void mkroot(int x) {
        access(x);
        splaying(x);
        reverse(x);
    }

    inline int findroot(int x) {
        access(x);
        splaying(x);
        while(ls(x)) {
            pushdown(x);
            x=ls(x);
        }
        splaying(x);
        return x;
    }

    inline void link(int x,int y) {
        mkroot(x);
        if(findroot(y)==x) return;
        fa(x)=y;
    }

    inline void cut(int x,int y) {
        mkroot(x);
        if(findroot(y)!=x||fa(y)!=x||ls(y)) return;
        fa(y)=rs(x)=0;
        pp(x);
    }

    inline void split(int x,int y) {
        mkroot(x);
        access(y);
        splaying(y);
    }
#undef fa
#undef ls
#undef rs
#undef ident
#undef reverse
#undef connect
#undef notroot
};

struct node{
    int op,x,y,id;
}q[N];

int ans[N],vis[N];
pair<int,int>e[N];
int n,m;
map<pair<int,int>,int>MP;

void add(int x,int y,int id){
    if(LCT::findroot(x) == LCT::findroot(y)){
        LCT::split(x,y);
        LCT::cover(y);
    }else{
        LCT::tree[id].val = 1;
        LCT::link(x,id);
        LCT::link(id,y);
    }
}

int query(int x,int y){
    LCT::split(x,y);
    return LCT::tree[y].res;
}

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif


    read(n,m);
    for(int i=1;i<=m;i++){
        int u,v;
        read(u,v);
        if(u>v) swap(u,v);
        e[i].fi = u;
        e[i].se = v;
        MP[e[i]] = i;
    }
    int cnt = 0;
    int op;
    while(read(op)){
        if(op==-1) break;
        int x,y;
        read(x,y);
        if(x>y) swap(x,y);
        q[++cnt] = {op,x,y,0};
        if(op==0){
            int id = MP[mp(x,y)];
            q[cnt].id = id;
            vis[id] = 1;
        }
    }

    for(int i=1;i<=m;i++){
        if(vis[i]) continue;
        add(e[i].fi,e[i].se,i+n);
    }

    for(int i=cnt;i>=1;i--){
        if(q[i].op) ans[i] = query(q[i].x,q[i].y);
        else add(q[i].x,q[i].y,q[i].id+n);
    }

    for(int i=1;i<=cnt;i++) if(q[i].op) printf("%d\n",ans[i]);

    return 0;
}