HDU 6200 边双联通分量 + 并查集 + dfs序 + BIT

简略题意：初始给出一张无向图，两种操作：
1. 添加一条(u,v)的无向边。
2. 问从u到v的路径上的割边有多少。

假若不考虑添加边的操作，问有多少割边，我们只需要边双联通缩个点成树，树上的每个边都是割边。从而转化成树上两点间距离。从根dfs一下转化成有根树的问题。

现在考虑添加边的过程，其实就是再缩点的过程，先不考虑如何缩点，假如缩了点之后，我们就需要动态更新两点间距离了。对此我们只需要用dfs序 + BIT维护每个点到根的距离即可。每次一个边(u−>v)如果失去了作用，那么以v为根的子树的值都需要−1。

考虑这个再缩点的过程，其实我们可以用并查集来维护，每个集合的根部点都代表了实际还存在的点。加入我们要把(u,v)相连，那么先找到lc=lca(u,v)，我们只需要暴力的把u和v向lc靠近，删除连接他们的边对答案的影响即可。

如果觉得我说的不是很清楚，可以参考叉姐的说法QAQ->ICPCCAMP。

#define others#ifdef poj#include <iostream>#include <cstring>#include <cmath>#include <cstdio>#include <algorithm>#include <vector>#include <string>#endif // poj#ifdef others#include <bits/stdc++.h>#endif // others//#define file#define all(x) x.begin(), x.end()using namespace std;const double eps = 1e-8;int dcmp(double x) { if(fabs(x)<=eps) return 0; return (x>0)?1:-1;};typedef long long LL;namespace fastIO{    #define BUF_SIZE 100000    #define OUT_SIZE 100000    #define ll long long    //fread->read    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;}            //{printf("IO error!\n");system("pause");for (;;);exit(0);}        }        return *p1++;    }    inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}    inline int read(int &x){        bool sign=0; char ch=nc(); x=0;        for (;blank(ch);ch=nc());        if (IOerror)return 0;        if (ch=='-')sign=1,ch=nc();        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';        if (sign)x=-x;        return 1;    }    inline int read(ll &x){        bool sign=0; char ch=nc(); x=0;        for (;blank(ch);ch=nc());        if (IOerror)return 0;        if (ch=='-')sign=1,ch=nc();        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';        if (sign)x=-x;        return 1;    }    inline int read(double &x){        bool sign=0; char ch=nc(); x=0;        for (;blank(ch);ch=nc());        if (IOerror)return 0;        if (ch=='-')sign=1,ch=nc();        for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0';        if (ch=='.'){            double tmp=1; ch=nc();            for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0');        }        if (sign)x=-x;        return 1;    }    inline int read(char *s){        char ch=nc();        for (;blank(ch);ch=nc());        if (IOerror)return 0;        for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch;        *s=0;        return 1;    }    inline void read(char &c){        for (c=nc();blank(c);c=nc());        if (IOerror){c=-1;return;}    }    //fwrite->write    struct Ostream_fwrite{        char *buf,*p1,*pend;        Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;}        void out(char ch){            if (p1==pend){                fwrite(buf,1,BUF_SIZE,stdout);p1=buf;            }            *p1++=ch;        }        void print(int x){            static char s[15],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1);        }        void println(int x){            static char s[15],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1); out('\n');        }        void print(ll x){            static char s[25],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1);        }        void println(ll x){            static char s[25],*s1;s1=s;            if (!x)*s1++='0';if (x<0)out('-'),x=-x;            while(x)*s1++=x%10+'0',x/=10;            while(s1--!=s)out(*s1); out('\n');        }        void print(double x,int y){            static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000,                1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL,                100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL};            if (x<-1e-12)out('-'),x=-x;x*=mul[y];            ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1;            ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2);            if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i); print(x3);}        }        void println(double x,int y){print(x,y);out('\n');}        void print(char *s){while (*s)out(*s++);}        void println(char *s){while (*s)out(*s++);out('\n');}        void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}}        ~Ostream_fwrite(){flush();}    }Ostream;    inline void print(int x){Ostream.print(x);}    inline void println(int x){Ostream.println(x);}    inline void print(char x){Ostream.out(x);}    inline void println(char x){Ostream.out(x);Ostream.out('\n');}    inline void print(ll x){Ostream.print(x);}    inline void println(ll x){Ostream.println(x);}    inline void print(double x,int y){Ostream.print(x,y);}    inline void println(double x,int y){Ostream.println(x,y);}    inline void print(char *s){Ostream.print(s);}    inline void println(char *s){Ostream.println(s);}    inline void println(){Ostream.out('\n');}    inline void flush(){Ostream.flush();}};using namespace fastIO;namespace solver {    const int maxn = 100011;    int n, m;    vector<int> G[maxn];    struct A {        int u, v, next;    } star[2*maxn];    stack<int> S;    int eg, head[maxn], dfn[maxn], low[maxn], towhere[maxn];    int Dindex, id;    bool instack[maxn];    void addedge(int u, int v) {        ++eg, star[eg] = {u, v, head[u]}, head[u] = eg;        swap(u, v);        ++eg, star[eg] = {u, v, head[u]}, head[u] = eg;    }    void dfsBCC(int u, int fa) {        dfn[u] = low[u] = ++Dindex;        S.push(u);        instack[u] = 1;        int v;        for(int i = head[u]; ~i; i = star[i].next) {            int v = star[i].v;            if(v == fa) continue;            if(!dfn[v]) {                dfsBCC(v, u);                low[u] = min(low[v], low[u]);            } else if(instack[v]) {                low[u] = min(low[u], dfn[v]);            }        }        if(low[u] == dfn[u]) {            id++;            for(;;) {                int tmp = S.top();                S.pop();                towhere[tmp] = id;                instack[tmp] = 0;                if(tmp == u) break;            }        }        return ;    }    void init() {        eg = Dindex = id = 0;        memset(head, -1, sizeof head);        while(!S.empty()) S.pop();        for(int i = 0; i < maxn; i++) G[i].clear();        memset(dfn, 0, sizeof dfn);        memset(instack, 0, sizeof instack);        memset(towhere, 0, sizeof towhere);    }    void rebuild() {        for(int i = 1; i <= eg; i+=2) {            int u = star[i].u, v = star[i].v;            u = towhere[u], v = towhere[v];            if(u == v) continue;            G[u].push_back(v);            G[v].push_back(u);        }    }    int time;    int C[maxn];    int lowbit(int x) {return x&-x;};    void add(int x, int v) { for(int i = x; i <= time; i+=lowbit(i)) C[i]+=v; };    int ask(int x) { int res = 0; for(int i = x; i; i -= lowbit(i)) res += C[i]; return res;};    int fa[maxn], L[maxn], R[maxn];    int find(int x) {        return x == fa[x]? x : fa[x] = find(fa[x]);    }    int p[20][maxn];    int deep[maxn];    void dfs(int u, int fa) {        p[0][u] = fa;        L[u] = ++time;        for(auto v : G[u])            if(v != fa) {                deep[v] = deep[u] + 1;                dfs(v, u);            }        R[u] = time;    }    int goup(int x, int len) {        for(int i = 19; i >= 0; i--)            if(x != -1 && (len & (1 << i)))                x = p[i][x];        return x;    }    int lca(int u, int v) {        if(deep[u] < deep[v]) swap(u, v);        int d = deep[u] - deep[v];        u = goup(u, d);        if(v == u) return u;        for(int i = 19; i >= 0; i--) {            if(p[i][u] == p[i][v]) continue;            u = p[i][u], v = p[i][v];        }        return p[0][u];    }    int solve() {        init();        read(n), read(m);        for(int i = 1; i <= m; i++) {            int u, v;            read(u), read(v);            addedge(u, v);        }        for(int i = 1; i <= n; i++)            if(!dfn[i]) dfsBCC(i, -1);        rebuild();        memset(C, 0, sizeof C);        time = 0;        deep[1] = 0;        dfs(1, -1);        for(int i = 1; i < 20; i++) {            for(int j = 1; j <= id; j++) {                if(p[i-1][j] == -1) p[i][j] = -1;                else p[i][j] = p[i-1][p[i-1][j]];            }        }        for(int i = 1; i <= id; i++) fa[i] = i, add(L[i], 1), add(R[i]+1, -1);        int q; read(q);        for(int i = 1; i <= q; i++) {            int pt, x, y;            read(pt); read(x); read(y);            x = towhere[x], y = towhere[y];            if(pt == 1) {                x = find(x), y = find(y);                if(x == y) continue;                int lc = lca(x, y);                lc = find(lc);                while(lc != x) {                    add(L[x], -1), add(R[x]+1, 1);                    fa[x] = lc;                    x = find(p[0][x]);                }                while(lc != y) {                    add(L[y], -1), add(R[y]+1, 1);                    fa[y] = lc;                    y = find(p[0][y]);                }            } else {                x = find(x), y = find(y);                int lc = lca(x, y);                lc = find(lc);                println(ask(L[x]) + ask(L[y]) - 2*ask(L[lc]));            }        }        return 0;    }}int main() {    int t;    scanf("%d", &t);    for(int i = 1; i <= t; i++) {        print("Case #"),print(i),print(":\n");        solver::solve();    }    return 0;}