【HDOJ】3686 Traffic Real Time Query System

这题做了几个小时，基本思路肯定是求两点路径中的割点数目，思路是tarjan缩点，然后以割点和连通块作为新节点见图。转化为lca求解。
结合点——双连通分量与LCA。

 /* 3686 */
 #include <iostream>
 #include <sstream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 typedef struct {
     int u, v, f, nxt;
 } edge_t;

 typedef struct {
     int v, nxt;
 } edge;

 const int maxv = ;
 const int maxe = ;
 int head[maxv], l, top;
 int pre[maxv], low[maxv];
 bool iscut[maxv];
 int cnt[maxv], dfs_clock, bcc_cnt;
 int bn[maxe];
 int S[maxe];
 vi bcc[maxv];
 edge_t E[maxe];
 int n, m;

 const int maxvv = maxv * ;
 int mark[maxvv];
 int head_[maxvv], l_;
 int cutn[maxvv];
 edge E_[maxe];

 int deep[maxvv], beg[maxvv];
 int V[maxvv<<], D[maxvv<<];

 int dp[][maxvv<<];

 void init_() {
     memset(head_, -, sizeof(head_));
     memset(mark, , sizeof(mark));
     l_ = ;
 }

 void addEdge_(int u, int v) {
     E_[l_].v = v;
     E_[l_].nxt = head_[u];
     head_[u] = l_++;

     E_[l_].v = u;
     E_[l_].nxt = head_[v];
     head_[v] = l_++;
 }

 void init() {
     l = dfs_clock = bcc_cnt = top = ;
     memset(head, -, sizeof(head));
     memset(iscut, false, sizeof(iscut));
     memset(cnt, , sizeof(cnt));
     memset(cutn, , sizeof(cutn));
     memset(pre, , sizeof(pre));
     rep(i, , n+)
         bcc[i].clr();
 }

 void addEdge(int u, int v) {
     E[l].u = u;
     E[l].f = ;
     E[l].v = v;
     E[l].nxt = head[u];
     head[u] = l++;

     E[l].u = v;
     E[l].f = ;
     E[l].v = u;
     E[l].nxt = head[v];
     head[v] = l++;
 }

 void tarjan(int u, int fa) {
     int v, k;

     low[u] = pre[u] = ++dfs_clock;
     for (k=head[u]; k!=-; k=E[k].nxt) {
         if (E[k].f)
             continue;
         E[k].f = E[k^].f = ;
         v = E[k].v;
         S[top++] = k;
         if (!pre[v]) {
             tarjan(v, u);
             low[u] = min(low[u], low[v]);
             if (low[v] >= pre[u]) {
                 iscut[u] = true;
                 ++cnt[u];
                 bcc_cnt++;
                 while () {
                     int kk = S[--top];
                     bn[kk>>] = bcc_cnt;
                     bcc[E[kk].u].pb(bcc_cnt);
                     bcc[E[kk].v].pb(bcc_cnt);
                     if (kk == k)
                         break;
                 }
             }
         } else {
             low[u] = min(low[u], pre[v]);
         }
     }
 }

 void dfs(int u, int fa, int d) {
     mark[u] = dfs_clock;
     deep[u] = d;
     V[++top] = u;
     D[top] = d;
     beg[u] = top;

     int v, k;

     for (k=head_[u]; k!=-; k=E_[k].nxt) {
         v = E_[k].v;
         if (v == fa)
             continue;
         dfs(v, u, d+);
         V[++top] = u;
         D[top] = d;
     }
 }

 void init_RMQ(int n) {
     int i, j;

     for (i=; i<=n; ++i)
         dp[][i] = i;
     for (j=; (<<j)<=n; ++j)
         for (i=; i+(<<j)-<=n; ++i)
             if (D[dp[j-][i]] < D[dp[j-][i+(<<(j-))]])
                 dp[j][i] = dp[j-][i];
             else
                 dp[j][i] = dp[j-][i+(<<(j-))];
 }

 int RMQ(int l, int r) {
     if (l > r)
         swap(l, r);

     int k = ;

     while (<<(k+) <= r-l+)
         ++k;

     if (D[dp[k][l]] < D[dp[k][r-(<<k)+]])
         return V[dp[k][l]];
     else
         return V[dp[k][r-(<<k)+]];
 }

 void solve() {
     int u, v, lca;

     rep(i, , n+) {
         if (!pre[i]) {
             tarjan(i, -);
             if (cnt[i] <= )
                 iscut[i] = false;
         }
     }

     int cid = bcc_cnt;

     init_();
     cid = bcc_cnt;
     for (u=; u<=n; ++u) {
         if (!iscut[u])
             continue;
         sort(all(bcc[u]));
         cutn[++cid] = ;
         addEdge_(cid, bcc[u][]);
         int sz = SZ(bcc[u]);
         rep(i, , sz) {
             if (bcc[u][i] != bcc[u][i-]) {
                 addEdge_(cid, bcc[u][i]);
             }
         }
     }

     top = ;
     ++dfs_clock;
     rep(i, , cid+) {
         if (mark[i] != dfs_clock)
             dfs(i, , );
     }

     init_RMQ(top);

     int q;
     int ans;

     scanf("%d", &q);
     while (q--) {
         scanf("%d %d", &u, &v);
         u = bn[u-];
         v = bn[v-];
         lca = RMQ(beg[u], beg[v]);
         ans = (deep[u]+deep[v] - deep[lca]* + ) / ;
         printf("%d\n", ans);
     }
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int u, v;

     while (scanf("%d %d", &n, &m)!=EOF && (n||m)) {
         init();
         rep(i, , m) {
             scanf("%d %d", &u, &v);
             addEdge(u, v);
         }

         solve();
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }