BZOJ 1969 航线规划

Solution

实际上就是查询 $u$ 到 $v$ 路径上 边双的个数 $ -1$。

并且题目仅有删边， 那么就离线倒序添边。

维护 边双 略有不同： 

首先需要一个并查集， 记录 边双内的点。 在 添加边$(u,v)$时 ， 若$u, v$ 已经相连， 那么把 $u, v$ 路径上的点 缩成一个点， 用最上面的点 来代替。

void del(int x, int y) {
        if (!x) return;
        fa[x] = y;
        del(lc(x), y);
        del(rc(x), y);
    }

access 也应该变化 ： $f[x]$ 可能已经经过压缩了。

void access(int x) {
        for (int y = 0; x; y = x, x = f[y] = anc(f[x]))
            splay(x), ch[x][1] = y, up(x);
    }

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rd read()
using namespace std;

const int N = 3e4 + 5;
const int M = 1e5 + 5;

int n, m;
int fa[N], ans[N], vis[M];

struct edge {
    int u, v;

    bool operator < (const edge &b) const {
        return u == b.u ? v < b.v : u < b.u;
    }
}e[M];

struct que {
    int op, u, v;
}q[M];

int read() {
    int X = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar())
        if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        X = X * 10 + c - '0';
    return X * p;
}

int anc(int x) {
    return fa[x] == x? x : fa[x] = anc(fa[x]);
}

namespace LCT {
    int f[N], sz[N], ch[N][2], tun[N];
#define lc(x) ch[x][0]
#define rc(x) ch[x][1]

    int isroot(int x) {
        return rc(f[x]) != x && lc(f[x]) != x;
    }

    int get(int x) {
        return rc(f[x]) == x;
    }

    void rev(int x) {
        swap(lc(x), rc(x));
        tun[x] ^= 1;
    }

    void up(int x) {
        sz[x] = 1;
        if (lc(x)) sz[x] += sz[lc(x)];
        if (rc(x)) sz[x] += sz[rc(x)];
    }

    void pushdown(int x) {
        if (tun[x]) {
            if (lc(x)) rev(lc(x));
            if (rc(x)) rev(rc(x));
            tun[x] = 0;
        }
    }

    void pd(int x) {
        if (!isroot(x))
            pd(f[x]);
        pushdown(x);
    }

    void rotate(int x) {
        int old = f[x], oldf = f[old], son = ch[x][get(x) ^ 1];
        if (!isroot(old)) ch[oldf][get(old)] = x;
        ch[x][get(x) ^ 1] = old;
        ch[old][get(x)] = son;
        f[son] = old; f[x] = oldf; f[old] = x;
        up(old); up(x);
    }

    void splay(int x) {
        pd(x);
        for (; !isroot(x); rotate(x))
            if (!isroot(f[x]))
                rotate(get(f[x]) == get(x) ? f[x] : x);
    }
    
    void access(int x) {
        for (int y = 0; x; y = x, x = f[y] = anc(f[x]))
            splay(x), ch[x][1] = y, up(x);
    }

    int findr(int x) {
        access(x); splay(x);
        while (lc(x)) pushdown(x), x = lc(x);
        return x;
    }

    void mroot(int x) {
        access(x); splay(x); rev(x);
    }

    void split(int x, int y) {
        mroot(x); access(y); splay(y);
    }

    void del(int x, int y) {
        if (!x) return;
        fa[x] = y;
        del(lc(x), y);
        del(rc(x), y);
    }

    void link(int x, int y) {
        mroot(x);
        f[x] = y;
    }

    void merge(int x, int y) {
        x = anc(x); y = anc(y);
        if (x == y)
            return;
        mroot(x);
        if (findr(y) != x) 
            link(x, y);
        else {
            splay(x);
            del(rc(x), x);
            rc(x) = 0; up(x);
        }
    }
}

using namespace LCT;

int main()
{
    n = rd; m = rd;
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
    for (int i = 1; i <= m; ++i) {
        e[i].u = rd; e[i].v = rd;
        if (e[i].u > e[i].v)
            swap(e[i].u, e[i].v);
    }
    sort(e + 1, e + 1 + m);
    int Q = 0;
    for (; ; ) {
        int op = rd;
        if (op == -1)
            break;
        int u = rd, v = rd;
        if (u > v)
            swap(u, v);
        q[++Q].op = op; q[Q].u = u;
        q[Q].v = v;
        edge t;
        t.u = u; t.v = v;
        if (op == 0)
            vis[lower_bound(e + 1, e + 1 + m, t) - e] = 1;
    }
    for (int i = 1; i <= m; ++i)
        if (!vis[i]) 
            merge(e[i].u, e[i].v);
    for (int i = Q; i; i--) {
        if (q[i].op == 1) {
            int x = q[i].u, y = q[i].v;
            x = anc(x); y = anc(y);
            if (x == y) {ans[i] = 0; continue;}
            split(x, y);
            ans[i] = sz[y] - 1;
        }
        else merge(q[i].u, q[i].v);
    }
    for (int i = 1; i <= Q; ++i)
        if (q[i].op == 1)
            printf("%d
", ans[i]);
}