POJ 3694 Network (算竞进阶习题)

边双联通分量

数据很水，随便乱搞。。

先把所有边双找出来，然后缩点。之后如果新加的边在同一个边双，那桥的数量不变，如果在不同的边双，就一直往上跳到两个点的LCA，通过的边如果没被标记，就标记一下，标记了就不再是桥了。。超级暴力！

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 200005;
int n, m, cnt, k, tot, t, ans, head[N], dfn[N], low[N], c[N], first[N], cur, p[N][20], depth[N], from[N];
bool bri[N<<2], kills[N<<2];
struct Edge { int v, next; } edge[N<<2], e[N<<2];

void addEdge(int a, int b){
    edge[cnt].v = b, edge[cnt].next = head[a], head[a] = cnt ++;
}

void link(int a, int b){
    e[cur].v = b, e[cur].next = first[a], first[a] = cur ++;
}

void tarjan(int s, int pre){
    dfn[s] = low[s] = ++k;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(!dfn[u]){
            tarjan(u, i);
            low[s] = min(low[s], low[u]);
            if(low[u] > dfn[s]) bri[i] = bri[i^1] = true;
        }
        else if(i != (pre^1)) low[s] = min(low[s], dfn[u]);
    }
}

void dfs(int s){
    c[s] = tot;
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(c[u] || bri[i]) continue;
        dfs(u);
    }
}

void dfs(int s, int fa){
    p[s][0] = fa, depth[s] = depth[fa] + 1;
    for(int i = 1; i <= t; i ++){
        p[s][i] = p[p[s][i - 1]][i - 1];
    }
    for(int i = first[s]; i != -1; i = e[i].next){
        int u = e[i].v;
        if(u == fa) continue;
        from[u] = i;
        dfs(u, s);
    }
}

int lca(int x, int y){
    if(depth[x] < depth[y]) swap(x, y);
    for(int i = t; i >= 0; i --){
        if(depth[p[x][i]] >= depth[y]) x = p[x][i];
    }
    if(x == y) return y;
    for(int i = t; i >= 0; i --){
        if(p[x][i] == p[y][i]) continue;
        x = p[x][i], y = p[y][i];
    }
    return p[y][0];
}

void build(){
    cnt = 2;
    k = tot = t = ans = cur = 0;
    full(head, -1), full(first, -1);
    full(dfn, 0), full(low, 0);
    full(c, 0), full(depth, 0);
    full(p, 0), full(from, 0);
    full(bri, false), full(kills, false);
}

int main(){

    int _ = 0;
    n = read(), m = read();
    while(n && m){
        build();
        for(int i = 0; i < m; i++){
            int u = read(), v = read();
            addEdge(u, v), addEdge(v, u);
        }
        for(int i = 1; i <= n; i++){
            if(!dfn[i]) tarjan(i, 0);
        }
        for(int i = 1; i <= n; i++){
            if(!c[i]) ++tot, dfs(i);
        }
        for(int i = 2; i < cnt; i += 2){
            int u = edge[i^1].v, v = edge[i].v;
            if(c[u] != c[v]) link(c[u], c[v]), link(c[v], c[u]);
            if(bri[i]) ans++;
        }
        t = (int) (log(n) / log(2)) + 1;
        dfs(1, 0);
        int q = read();
        printf("Case %d:\n", ++_);
        while(q--){
            int u = read(), v = read();
            if(c[u] == c[v]) printf("%d\n", ans);
            else{
                int num = 0;
                int f = lca(c[u], c[v]);
                int tmp = c[u];
                while(tmp != f){
                    if(!kills[from[tmp]]){
                        kills[from[tmp]] = kills[from[tmp]^1] = true;
                        num++;
                    }
                    tmp = p[tmp][0];
                }
                tmp = c[v];
                while(tmp != f){
                    if(!kills[from[tmp]]){
                        kills[from[tmp]] = kills[from[tmp]^1] = true;
                        num++;
                    }
                    tmp = p[tmp][0];
                }
                ans -= num;
                printf("%d\n", ans);
            }
        }
        n = read(), m = read();
    }
    return 0;
}