并查集+树链剖分+线段树 HDOJ 5458 Stability（稳定性）

题目链接

题意：

　　有n个点m条边的无向图，有环还有重边，a到b的稳定性的定义是有多少条边，单独删去会使a和b不连通。有两种操作：

　　1. 删去a到b的一条边

　　2. 询问a到b的稳定性

思路：

　　首先删边考虑离线，倒着做，相对于加边。先用并查集建一棵树，最精简的图，初始化树上的每条边权值为1，那么在a和b点加一条边的话，会使a到b的链上所有边因为这条新边而稳定性贡献无效，这样我们可以树链剖分，用线段树维护链上的边的稳定性贡献值。

#include <bits/stdc++.h>
using namespace std;

const int N = 3e4 + 5;
const int M = 1e5 + 5;
int n, m, q;

struct Edge {
    int v, nex;
}edge[M<<2];
int head[N];
int etot;

void init_edge() {
    memset (head, -1, sizeof (head));
    etot = 0;
}

void add_edge(int u, int v) {
    edge[etot] = (Edge) {v, head[u]};
    head[u] = etot++;
}

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1

int sum[N<<2], lazy[N<<2];

void push_up(int o) {
    sum[o] = sum[o<<1] + sum[o<<1|1];
}

void push_down(int l, int r, int o) {
    if (lazy[o] != -1) {
        lazy[o<<1] = lazy[o];
        lazy[o<<1|1] = lazy[o];
        int mid = l + r >> 1;
        sum[o<<1] = lazy[o] * (mid - l + 1);
        sum[o<<1|1] = lazy[o] * (r - mid + 1);
        lazy[o] = -1;
    }
}

void build(int l, int r, int o) {
    lazy[o] = -1;
    if (l == r) {
        sum[o] = 1;
        return ;
    }
    int mid = l + r >> 1;
    build (lson);
    build (rson);
    push_up (o);
}

int query(int ql, int qr, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        return sum[o];
    }
    push_down (l, r, o);
    int mid = l + r >> 1, ret = 0;
    if (ql <= mid) ret += query (ql, qr, lson);
    if (qr > mid) ret += query (ql, qr, rson);
    return ret;
}

void updata(int ql, int qr, int c, int l, int r, int o) {
    if (ql <= l && r <= qr) {
        sum[o] = (r - l + 1) * c;
        lazy[o] = c;
        return ;
    }
    push_down (l, r, o);
    int mid = l + r >> 1;
    if (ql <= mid) updata (ql, qr, c, lson);
    if (qr > mid) updata (ql, qr, c, rson);
    push_up (o);
}

int sz[N], fa[N], dfn[N], belong[N];
int tim;

void DFS2(int u, int chain) {
    dfn[u] = ++tim;
    belong[u] = chain;
    int k = 0;
    for (int i=head[u]; ~i; i=edge[i].nex) {
        Edge &e = edge[i];
        if (e.v == fa[u]) continue;
        if (sz[e.v] > sz[k]) k = e.v;
    }
    if (k) DFS2 (k, chain);
    for (int i=head[u]; ~i; i=edge[i].nex) {
        Edge &e = edge[i];
        if (e.v == fa[u] || e.v == k) continue;
        DFS2 (e.v, e.v);
    }
}

void DFS(int u, int pa) {
    sz[u] = 1;
    fa[u] = pa;
    for (int i=head[u]; ~i; i=edge[i].nex) {
        Edge &e = edge[i];
        if (e.v == pa) continue;
        DFS (e.v, u);
        sz[u] += sz[e.v];
    }
}

int query_ans(int u, int v) {
    int ret = 0;
    int p = belong[u], q = belong[v];
    while (p != q) {
        if (dfn[p] < dfn[q]) {
            swap (p, q);
            swap (u, v);
        }
        ret += query (dfn[p], dfn[u], 1, n, 1);
        u = fa[p];
        p = belong[u];
    }
    if (dfn[u] < dfn[v]) swap (u, v);
    if (u != v) {
        ret += query (dfn[v]+1, dfn[u], 1, n, 1);
    }
    return ret;
}

void modify(int u, int v) {
    int p = belong[u], q = belong[v];
    while (p != q) {
        if (dfn[p] < dfn[q]) {
            swap (p, q);
            swap (u, v);
        }
        updata (dfn[p], dfn[u], 0, 1, n, 1);
        u = fa[p];
        p = belong[u];
    }
    if (dfn[u] < dfn[v]) swap (u, v);
    if (u != v) {
        updata (dfn[v]+1, dfn[u], 0, 1, n, 1);
    }
}

void prepare() {
    sz[0] = 0;
    DFS (1, 0);
    tim = 0;
    DFS2 (1, 1);
    build (1, n, 1);
}

int rt[N];

int Find(int x) {
    return rt[x] == x ? x : rt[x] = Find (rt[x]);
}

multiset<pair<int, int> > mset1, mset2;
int op[M], x[M], y[M];
int ans[M];

int main() {
    int T, cas = 0;
    scanf ("%d", &T);
    while (T--) {
        init_edge ();
        scanf ("%d%d%d", &n, &m, &q);
        for (int i=1; i<=n; ++i) {
            rt[i] = i;
        }
        mset1.clear ();
        for (int i=1; i<=m; ++i) {
            int u, v;
            scanf ("%d%d", &u, &v);
            if (u > v) swap (u, v);
            mset1.insert ({u, v});
        }
        for (int i=1; i<=q; ++i) {
            scanf ("%d%d%d", &op[i], &x[i], &y[i]);
            if (x[i] > y[i]) swap (x[i], y[i]);
            if (op[i] == 1) mset1.erase (mset1.find ({x[i], y[i]}));
        }
        mset2.clear ();
        for (auto &it: mset1) {
            int p = Find (it.first), q = Find (it.second);
            if (p == q) continue;
            mset2.insert (it);
            add_edge (it.first, it.second);
            add_edge (it.second, it.first);
            rt[p] = q;
        }

        prepare ();

        for (auto &it: mset1) {
            if (mset2.find (it) == mset2.end ()) {
                modify (it.first, it.second);
            }
        }
        for (int i=q; i>=1; --i) {
            if (op[i] == 1) {
                modify (x[i], y[i]);
            } else {
                ans[i] = query_ans (x[i], y[i]);
            }
        }

        printf ("Case #%d:\n", ++cas);
        for (int i=1; i<=q; ++i) {
            if (op[i] == 2) {
                printf ("%d\n", ans[i]);
            }
        }
    }
    return 0;
}