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 ith line (i starts from 1)contains two integers Xi and Yi, representing that roadi connects crossing Xi and Yi (Xi≠Yi).
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<Xi,Yi<=N, 0<S,T<=M
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 ith line (i starts from 1)contains two integers Xi and Yi, representing that roadi connects crossing Xi and Yi (Xi≠Yi).
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<Xi,Yi<=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):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <cstdio> 2 #include <iostream> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 using namespace std; 7 typedef long long LL; 8 9 typedef pair<int, int> PII; 10 11 const int MAXV = 10010; 12 const int MAXE = 200010; 13 14 int ans[MAXV]; 15 vector<PII> query[MAXV << 1]; 16 17 struct SccGraph { 18 int head[MAXV << 1], fa[MAXV << 1], ecnt; 19 bool vis[MAXV << 1]; 20 int to[MAXE << 1], next[MAXE << 1]; 21 int dep[MAXV << 1]; 22 23 void init(int n) { 24 memset(head, -1, sizeof(int) * (n + 1)); 25 memset(vis, 0, sizeof(bool) * (n + 1)); 26 for(int i = 1; i <= n; ++i) fa[i] = i; 27 ecnt = 0; 28 } 29 30 int find_set(int x) { 31 return fa[x] == x ? x : fa[x] = find_set(fa[x]); 32 } 33 34 void add_edge(int u, int v) { 35 to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++; 36 to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++; 37 } 38 39 void lca(int u, int f, int deep) { 40 dep[u] = deep; 41 for(int p = head[u]; ~p; p = next[p]) { 42 int &v = to[p]; 43 if(v == f || vis[v]) continue; 44 lca(v, u, deep + 1); 45 fa[v] = u; 46 } 47 vis[u] = true; 48 for(vector<PII>::iterator it = query[u].begin(); it != query[u].end(); ++it) { 49 if(vis[it->first]) { 50 ans[it->second] = (dep[u] + dep[it->first] - 2 * dep[find_set(it->first)]) / 2; 51 } 52 } 53 } 54 } G; 55 56 int head[MAXV], lowlink[MAXV], pre[MAXV], ecnt, dfs_clock; 57 int sccno[MAXV], scc_cnt; 58 int to[MAXE], next[MAXE], scc_edge[MAXE]; 59 bool vis[MAXE], iscut[MAXV]; 60 int stk[MAXE], top; 61 int n, m, q; 62 63 void init() { 64 memset(head, -1, sizeof(int) * (n + 1)); 65 memset(pre, 0, sizeof(int) * (n + 1)); 66 memset(iscut, 0, sizeof(bool) * (n + 1)); 67 memset(vis, 0, sizeof(bool) * (2 * m)); 68 ecnt = scc_cnt = dfs_clock = 0; 69 } 70 71 void add_edge(int u, int v) { 72 to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++; 73 to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++; 74 } 75 76 void tarjan(int u, int f) { 77 pre[u] = lowlink[u] = ++dfs_clock; 78 int child = 0; 79 for(int p = head[u]; ~p; p = next[p]) { 80 if(vis[p]) continue; 81 vis[p] = vis[p ^ 1] = true; 82 stk[++top] = p; 83 int &v = to[p]; 84 if(!pre[v]) { 85 ++child; 86 tarjan(v, u); 87 lowlink[u] = min(lowlink[u], lowlink[v]);\ 88 if(pre[u] <= lowlink[v]) { 89 iscut[u] = true; 90 ++scc_cnt; 91 while(true) { 92 int t = stk[top--]; 93 scc_edge[t] = scc_edge[t ^ 1] = scc_cnt; 94 if(t == p) break; 95 } 96 } 97 } else lowlink[u] = min(lowlink[u], pre[v]); 98 } 99 if(f < 1 && child == 1) iscut[u] = false; 100 } 101 102 void build() { 103 G.init(scc_cnt); 104 for(int p = 0; p != ecnt; ++p) { 105 int &v = to[p]; 106 if(iscut[v]) G.add_edge(sccno[v], scc_edge[p]); 107 } 108 } 109 110 void solve() { 111 for(int i = 1; i <= n; ++i) 112 if(!pre[i]) tarjan(i, 0); 113 for(int u = 1; u <= n; ++u) 114 if(iscut[u]) sccno[u] = ++scc_cnt; 115 } 116 117 int main() { 118 while(scanf("%d%d", &n, &m) != EOF) { 119 if(n == 0 && m == 0) break; 120 init(); 121 for(int i = 1; i <= m; ++i) { 122 int u, v; 123 scanf("%d%d", &u, &v); 124 add_edge(u, v); 125 } 126 solve(); 127 build(); 128 for(int i = 0; i <= scc_cnt; ++i) query[i].clear(); 129 scanf("%d", &q); 130 for(int i = 0; i < q; ++i) { 131 int x, y; 132 scanf("%d%d", &x, &y); 133 x = scc_edge[x * 2 - 2]; y = scc_edge[y * 2 - 2]; 134 query[x].push_back(make_pair(y, i)); 135 query[y].push_back(make_pair(x, i)); 136 } 137 for(int i = 1; i <= scc_cnt; ++i) if(!G.vis[i]) G.lca(i, 0, 0); 138 for(int i = 0; i < q; ++i) printf("%d\n", ans[i]); 139 } 140 }