CF 1051 F. The Shortest Statement

F. The Shortest Statement

http://codeforces.com/contest/1051/problem/F

题意：

　　n个点，m条边的无向图，每次询问两点之间的最短路。（m-n<=20）

分析：

　　dijkstra。

　　如果是一棵树，那么可以直接通过，dis[u]+dis[v]-dis[lca]*2来求。现在如果建出一棵树，那么非树边只有小于等于21条。

　　只经过树边的路径用上面的方式求出，考虑经过非树边的路径。

　　经过非树边（至少一条），那么一定经过了这条边的顶点，所以可以对顶点做一次最短路，如果询问u,v经过这个顶点，那么就是dis[u]+dis[v]。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<cctype>
 #include<set>
 #include<vector>
 #include<queue>
 #include<map>
 #define fi(s) freopen(s,"r",stdin);
 #define fo(s) freopen(s,"w",stdout);
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const LL INF = 1e18;
 const int N = ;

 int head[N], nxt[N << ], to[N << ], len[N << ], En;
 int f[N][], deth[N], tmp[], tot;
 LL dis[N], d[][N];
 int n, m;
 bool vis[N];

 void add_edge(int u,int v,int w) {
     ++En, to[En] = v, len[En] = w, nxt[En] = head[u], head[u] = En;
     ++En, to[En] = u, len[En] = w, nxt[En] = head[v], head[v] = En;
 }

 #define pa pair<LL,int>
 #define mp(a,b) make_pair(a,b)
 priority_queue< pa, vector< pa >, greater< pa > > q;

 void Dijkstra(int id,int S) {
     for (int i=; i<=n; ++i) dis[i] = INF, vis[i] = false;
     dis[S] = ;
     q.push(mp(, S));
     while (!q.empty()) {
         int u = q.top().second; q.pop();
         if (vis[u]) continue;
         vis[u] = true;
         for (int i=head[u]; i; i=nxt[i]) {
             int v = to[i];
             if (dis[v] > dis[u] + len[i]) {
                 dis[v] = dis[u] + len[i];
                 q.push(mp(dis[v], v));
             }
         }
     }
     for (int i=; i<=n; ++i) d[id][i] = dis[i];
 }
 void dfs(int u,int fa) {
     vis[u] = true;
     f[u][] = fa;
     deth[u] = deth[fa] + ;
     for (int i=head[u]; i; i=nxt[i]) {
         int v = to[i];
         if (v == fa) continue;
         if (vis[v]) tmp[++tot] = u, tmp[++tot] = v;
         else dis[v] = dis[u] + len[i], dfs(v, u);
     }
 }
 int LCA(int u,int v) {
     if (deth[u] < deth[v]) swap(u, v);
     int d = deth[u] - deth[v];
     for (int i=; i>=; --i)
         if (d & ( << i)) u = f[u][i];
     if (u == v) return u;
     for (int i=; i>=; --i)
         if (f[u][i] != f[v][i])
             u = f[u][i], v = f[v][i];
     return f[u][];
 }
 int main() {
     n = read(), m = read();
     for (int i=; i<=m; ++i) {
         int u = read(), v = read(), w = read();
         add_edge(u, v, w);
     }

     dfs(, );
     for (int i=; i<=n; ++i) d[][i] = dis[i];
     for (int j=; j<=; ++j)
         for (int i=; i<=n; ++i) f[i][j] = f[f[i][j-]][j-];
     sort(tmp + , tmp + tot + );
     int lim = tot; tot = ;
     for (int i=; i<=lim; ++i) if (tmp[tot] != tmp[i]) tmp[++tot] = tmp[i];
     for (int i=; i<=tot; ++i) Dijkstra(i, tmp[i]);

     int Q = read();
     while (Q --) {
         int u = read(), v = read();
         int t = LCA(u, v);
         LL ans = d[][u] + d[][v] -  * d[][t];
         for (int i=; i<=tot; ++i)
             ans = min(ans, d[i][u] + d[i][v]);
         printf("%I64d\n",ans);
     }
     return ;
 }