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;
- }
来源: http://www.bubuko.com/infodetail-3051534.html