HDU 3686 Traffic Real Time Query System（双连通分量缩点+LCA）（2010 Asia Hangzhou Regional Contest）

Problem Description

City C is really a nightmare of all drivers for its traffic jams. To solve the traffic problem, the mayor plans to build a RTQS (Real Time Query System) to monitor all traffic situations. City C is made up of N crossings and M roads, and each road connects two crossings. All roads are bidirectional. One of the important tasks of RTQS is to answer some queries about route-choice problem. Specifically, the task is to find the crossings which a driver MUST pass when he is driving from one given road to another given road.

 

Input

There are multiple test cases.
For each test case:
The first line contains two integers N and M, representing the number of the crossings and roads.
The next M lines describe the roads. In those M lines, the i^th line (i starts from 1)contains two integers X_i and Y_i, representing that road_i connects crossing X_i and Y_i (X_i≠Y_i).
The following line contains a single integer Q, representing the number of RTQs.
Then Q lines follows, each describing a RTQ by two integers S and T(S≠T) meaning that a driver is now driving on the roads and he wants to reach roadt . It will be always at least one way from roads to roadt.
The input ends with a line of “0 0”.
Please note that: 0<N<=10000, 0<M<=100000, 0<Q<=10000, 0<X_i,Y_i<=N, 0<S,T<=M 

 

Output

For each RTQ prints a line containing a single integer representing the number of crossings which the driver MUST pass.

 

题目大意：给一个N个点M条边的无向图，然后有Q个询问X，Y，问第X条边到第Y条边必需要经过的点有多少个。

思路：对于两条边X，Y，若他们在同一个双连通分量中，那他们之间至少有两条路径可以互相到达。

那么，对于两条不在同一个分量中的边X，Y，显然从X到Y必须要经过的的点的数目等于从X所在的边双连通分量到Y所在的双连通分量中的割点的数目。

于是，可以找出所有双连通分量，缩成一个点。对于 双连通分量A - 割点 - 双连通分量B，重构图成3个点 A - 割点 - B。

那么，所有的双连通分量缩完之后，新图G‘成了一棵树（若存在环，那么环里的点都在同一个双连通分量中，矛盾）

那么题目变成了：从X所在的G'中的点，到Y所在的G’中的点的路径中，有多少点是由割点变成的。

由于图G‘中的点都是 双连通分量 - 割点 - 双连通分量 - 割点 - 双连通分量……的形式（两个双连通分量不会连在一起）

那么X到Y的割点的数目就等于两点距离除以2，暨(dep[x] + dep[Y] - dep[LCA(X, Y)]) / 2，其中dep是深度，lca是最近公共祖先。

其中LCA用tarjan也好，用RMQ也好，随意。我用的是离线的tarjan。

 

细节：因为找边双连通分量的思路错了跪了一整天……写一下正确又优美的姿势是怎么样的>_<

类似于tarjan求强联通的算法，用一个vis[]数组记录某条边是否被访问过。

每次第一次访问一条边，把这条边压入栈中，在遍历完某条边指向的点之后，若pre[u] <= lowlink[v]（见code），把栈中的边都设为同一个边双连通分量。

仔细想想觉得应该是对的。我不会证明>_<

 

PS：新图G’中点的数目可能会高达2*n的等级，试想一下原图恰好是一条链的情况，由于存在重边，边数也最好要2*m。

 

代码（178MS）：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;
 typedef long long LL;

 typedef pair<int, int> PII;

 const int MAXV = ;
 const int MAXE = ;

 int ans[MAXV];
 vector<PII> query[MAXV << ];

 struct SccGraph {
     int head[MAXV << ], fa[MAXV << ], ecnt;
     bool vis[MAXV << ];
     int to[MAXE << ], next[MAXE << ];
     int dep[MAXV << ];

     void init(int n) {
         memset(head, -, sizeof(int) * (n + ));
         memset(vis, , sizeof(bool) * (n + ));
         for(int i = ; i <= n; ++i) fa[i] = i;
         ecnt = ;
     }

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

     void add_edge(int u, int v) {
         to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
         to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
     }

     void lca(int u, int f, int deep) {
         dep[u] = deep;
         for(int p = head[u]; ~p; p = next[p]) {
             int &v = to[p];
             if(v == f || vis[v]) continue;
             lca(v, u, deep + );
             fa[v] = u;
         }
         vis[u] = true;
         for(vector<PII>::iterator it = query[u].begin(); it != query[u].end(); ++it) {
             if(vis[it->first]) {
                 ans[it->second] = (dep[u] + dep[it->first] -  * dep[find_set(it->first)]) / ;
             }
         }
     }
 } G;

 int head[MAXV], lowlink[MAXV], pre[MAXV], ecnt, dfs_clock;
 int sccno[MAXV], scc_cnt;
 int to[MAXE], next[MAXE], scc_edge[MAXE];
 bool vis[MAXE], iscut[MAXV];
 int stk[MAXE], top;
 int n, m, q;

 void init() {
     memset(head, -, sizeof(int) * (n + ));
     memset(pre, , sizeof(int) * (n + ));
     memset(iscut, , sizeof(bool) * (n + ));
     memset(vis, , sizeof(bool) * ( * m));
     ecnt = scc_cnt = dfs_clock = ;
 }

 void add_edge(int u, int v) {
     to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
     to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
 }

 void tarjan(int u, int f) {
     pre[u] = lowlink[u] = ++dfs_clock;
     int child = ;
     for(int p = head[u]; ~p; p = next[p]) {
         if(vis[p]) continue;
         vis[p] = vis[p ^ ] = true;
         stk[++top] = p;
         int &v = to[p];
         if(!pre[v]) {
             ++child;
             tarjan(v, u);
             lowlink[u] = min(lowlink[u], lowlink[v]);\
             if(pre[u] <= lowlink[v]) {
                 iscut[u] = true;
                 ++scc_cnt;
                 while(true) {
                     int t = stk[top--];
                     scc_edge[t] = scc_edge[t ^ ] = scc_cnt;
                     if(t == p) break;
                 }
             }
         } else lowlink[u] = min(lowlink[u], pre[v]);
     }
     if(f <  && child == ) iscut[u] = false;
 }

 void build() {
     G.init(scc_cnt);
     for(int p = ; p != ecnt; ++p) {
         int &v = to[p];
         if(iscut[v]) G.add_edge(sccno[v], scc_edge[p]);
     }
 }

 void solve() {
     for(int i = ; i <= n; ++i)
         if(!pre[i]) tarjan(i, );
     for(int u = ; u <= n; ++u)
         if(iscut[u]) sccno[u] = ++scc_cnt;
 }

 int main() {
     while(scanf("%d%d", &n, &m) != EOF) {
         if(n ==  && m == ) break;
         init();
         for(int i = ; i <= m; ++i) {
             int u, v;
             scanf("%d%d", &u, &v);
             add_edge(u, v);
         }
         solve();
         build();
         for(int i = ; i <= scc_cnt; ++i) query[i].clear();
         scanf("%d", &q);
         for(int i = ; i < q; ++i) {
             int x, y;
             scanf("%d%d", &x, &y);
             x = scc_edge[x *  - ]; y = scc_edge[y *  - ];
             query[x].push_back(make_pair(y, i));
             query[y].push_back(make_pair(x, i));
         }
         for(int i = ; i <= scc_cnt; ++i) if(!G.vis[i]) G.lca(i, , );
         for(int i = ; i < q; ++i) printf("%d\n", ans[i]);
     }
 }