ZOJ - 3195 Design the city

题目要对每次询问将一个树形图的三个点连接，输出最短距离。

利用tarjan离线算法，算出每次询问的任意两个点的最短公共祖先，并在dfs过程中求出离根的距离。把每次询问的三个点两两求出最短距离，这样最终结果就是3个值一半。

其实开始我用的一种很挫的方法才AC的，具体思路就不说了，感觉很麻烦又不好写的样子。怎么没想到上面的简便方法呢。

初始代码：

 #include <iostream>
 #include <sstream>
 #include <cstdio>
 #include <climits>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <stack>
 #include <map>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <algorithm>
 #define esp 1e-6
 #define pi acos(-1.0)
 #define pb push_back
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define mp(a, b) make_pair((a), (b))
 #define in  freopen("in.txt", "r", stdin);
 #define out freopen("out.txt", "w", stdout);
 #define print(a) printf("%d\n",(a));
 #define bug puts("********))))))");
 #define stop  system("pause");
 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
 #define inf 0x0f0f0f0f

 using namespace std;
 typedef long long  LL;
 typedef vector<int> VI;
 typedef pair<int, int> pii;
 typedef vector<pii> VII;
 typedef vector<pii, int> VIII;
 typedef VI:: iterator IT;
 const int maxn =  + ;
 const int maxm = ( + ) * ;
 int dis[maxn], lin[maxm][], vis[maxn], pa[maxn];
 VII g[maxn];
 VII query[maxn];
 int n, m;
 int findset(int x)
 {
     return pa[x] == x? x : pa[x] = findset(pa[x]);
 }
 void tarjan(int u)
 {
     vis[u] = ;
     pa[u] = u;
     for(int i = ; i < (int)query[u].size(); i++)
     {
         int v= query[u][i].first;
         if(vis[v])
         {
             lin[query[u][i].second][] = findset(v);
         }
     }
     for(int i = ; i < (int)g[u].size(); i++)
     {
         int v = g[u][i].second;
         if(!vis[v])
         {
             dis[v] = dis[u] + g[u][i].first;
             tarjan(v);
             pa[v] = u;
         }
     }
 }
 void Init(void)
 {
     for(int i = ; i < maxn; i++)
         query[i].clear(), g[i].clear();
     memset(vis, , sizeof(vis));
 }
 int main(void)
 {
     int flag = ;
     while(scanf("%d", &n) == )
     {
         if(flag) puts("");
         else flag = ;
         Init();
         for(int i = ; i < n; i++)
         {
             int u, v, len;
             scanf("%d%d%d", &u, &v, &len);
             g[u].pb(mp(len, v));
             g[v].pb(mp(len, u));
         }
         scanf("%d", &m);
         for(int i = ; i <= *m; i += )
         {
             int x, y, z;
             scanf("%d%d%d", &x, &y, &z);
             query[lin[i][] = x].pb(mp(lin[i][] = y, i));
             query[lin[i][]].pb(mp(lin[i][], i));
             query[lin[i+][] = x].pb(mp(lin[i+][] = z, i+));
             query[lin[i+][]].pb(mp(lin[i+][], i+));
             query[lin[i+][] = y].pb(mp(lin[i+][] = z, i+));
             query[lin[i+][]].pb(mp(lin[i+][], i+));
         }
         dis[] = ;
         tarjan();
         for(int i = ; i <= *m; i += )
         {
             int ans;
             if(lin[i+][] ==  lin[i+][])
             {
 //                if(lin[i+2][2] == 0)
 //                ans = dis[lin[i][0]] + dis[lin[i][1]] - 2 * dis[lin[i][2]] + dis[lin[i][2]] + dis[lin[i+1][1]];
                 ans = dis[lin[i][]] + dis[lin[i][]] -  * dis[lin[i][]] - *dis[lin[i+][]] + dis[lin[i+][]] + dis[lin[i][]];
             }
             else
                 ans = dis[lin[i][]] + dis[lin[i][]] -  * dis[lin[i][]] + dis[lin[i+][]]- max(dis[lin[i+][]], dis[lin[i+][]]);
             printf("%d\n",ans);
         }
     }
     return ;
 }

简便方法的代码：

 #include <iostream>
 #include <sstream>
 #include <cstdio>
 #include <climits>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <stack>
 #include <map>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <algorithm>
 #define esp 1e-6
 #define pi acos(-1.0)
 #define pb push_back
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 #define mp(a, b) make_pair((a), (b))
 #define in  freopen("in.txt", "r", stdin);
 #define out freopen("out.txt", "w", stdout);
 #define print(a) printf("%d\n",(a));
 #define bug puts("********))))))");
 #define stop  system("pause");
 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
 #define inf 0x0f0f0f0f

 using namespace std;
 typedef long long  LL;
 typedef vector<int> VI;
 typedef pair<int, int> pii;
 typedef vector<pii> VII;
 typedef vector<pii, int> VIII;
 typedef VI:: iterator IT;
 const int maxn =  + ;
 const int maxm = ( + ) * ;
 int dis[maxn], lin[maxm][], vis[maxn], pa[maxn];
 VII g[maxn];
 VII query[maxn];
 int ans[maxm];
 int n, m;
 int findset(int x)
 {
     return pa[x] == x? x : pa[x] = findset(pa[x]);
 }
 void tarjan(int u)
 {
     vis[u] = ;
     pa[u] = u;
     for(int i = ; i < (int)query[u].size(); i++)
     {
         int v= query[u][i].first;
         if(vis[v])
         {
             ans[query[u][i].second] += dis[u] + dis[v] -  * dis[findset(v)];
         }
     }
     for(int i = ; i < (int)g[u].size(); i++)
     {
         int v = g[u][i].second;
         if(!vis[v])
         {
             dis[v] = dis[u] + g[u][i].first;
             tarjan(v);
             pa[v] = u;
         }
     }
 }
 void Init(void)
 {
     for(int i = ; i < maxn; i++)
         query[i].clear(), g[i].clear();
     memset(vis, , sizeof(vis));
     memset(ans, , sizeof(ans));
 }
 int main(void)
 {
     int flag = ;
     while(scanf("%d", &n) == )
     {
         if(flag) puts("");
         else flag = ;
         Init();
         for(int i = ; i < n; i++)
         {
             int u, v, len;
             scanf("%d%d%d", &u, &v, &len);
             g[u].pb(mp(len, v));
             g[v].pb(mp(len, u));
         }
         scanf("%d", &m);
         for(int i = ; i <= m; i++)
         {
             int x, y, z;
             scanf("%d%d%d", &x, &y, &z);
             query[x].pb(mp(y, i));
             query[y].pb(mp(x, i));
             query[x].pb(mp(z, i));
             query[z].pb(mp(x, i));
             query[y].pb(mp(z, i));
             query[z].pb(mp(y, i));
         }
         dis[] = ;
         tarjan();
         for(int i = ; i <= m; i++)
         {
             printf("%d\n", ans[i]>>);
         }
     }
     return ;
 }