【CF】121 Div.1 C. Fools and Roads

题意是给定一棵树。同时，给定如下k个查询：

给出任意两点u,v，对u到v的路径所经过的边进行加计数。

k个查询后，分别输出各边的计数之和。

思路利用LCA，对cnt[u]++, cnt[v]++,并对cnt[LCA(u, v)] -= 2.
然后dfs求解各边的计数。

 /* 191C */
 #include <iostream>
 #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

 const int maxn = 1e5+;
 const int maxd = ;
 vpii E[maxn];
 bool visit[maxn];
 int ans[maxn];
 int cnt[maxn];
 int deep[maxn];
 int  pre[maxn][maxd];

 void dfs(int u, int fa) {
     int sz = SZ(E[u]), v;

     deep[u] = deep[fa] + ;
     rep(i, , sz) {
         v = E[u][i].fir;
         if (v == fa)
             continue;
         pre[v][] = u;
         rep(j, , maxd)
             pre[v][j] = pre[pre[v][j-]][j-];
         dfs(v, u);
     }
 }

 int LCA(int u, int v)  {
     if (deep[u] > deep[v])
         swap(u, v);
     if (deep[u] < deep[v]) {
         int tmp = deep[v] - deep[u];
         rep(i, , maxd) {
             if (tmp & (<<i))
                 v = pre[v][i];
         }
     }

     if (u != v) {
         per(i, , maxd) {
             if (pre[u][i] != pre[v][i]) {
                 u = pre[u][i];
                 v = pre[v][i];
             }
         }
         return pre[u][];
     } else {
         return u;
     }
 }

 int dfs2(int u, int eid) {
     int ret = , sz = SZ(E[u]);
     int e, v;

     visit[u] = true;
     rep(i, , sz) {
         v = E[u][i].fir;
         e = E[u][i].sec;
         if (!visit[v]) {
             ret += dfs2(v, e);
         }
     }
     ret += cnt[u];
     ans[eid] = ret;

     return ret;
 }

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

     int n;
     int u, v, fa;

     scanf("%d", &n);
     rep(i, , n) {
         scanf("%d %d", &u, &v);
         E[u].pb(mp(v, i));
         E[v].pb(mp(u, i));
     }

     dfs(, );

     int k;

     scanf("%d", &k);
     while (k--) {
         scanf("%d %d", &u, &v);
         fa = LCA(u, v);
         ++cnt[u];
         ++cnt[v];
         cnt[fa] -= ;
     }

     dfs2(, );
     rep(i, , n)
         printf("%d ", ans[i]);
     putchar('\n');

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

     return ;
 }