【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+5;
const int maxd = 17;
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] + 1;
rep(i, 0, sz) {
v = E[u][i].fir;
if (v == fa)
continue;
pre[v][0] = u;
rep(j, 1, maxd)
pre[v][j] = pre[pre[v][j-1]][j-1];
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, 0, maxd) {
if (tmp & (1<<i))
v = pre[v][i];
}
}

if (u != v) {
per(i, 0, maxd) {
if (pre[u][i] != pre[v][i]) {
u = pre[u][i];
v = pre[v][i];
}
}
return pre[u][0];
} else {
return u;
}
}

int dfs2(int u, int eid) {
int ret = 0, sz = SZ(E[u]);
int e, v;

visit[u] = true;
rep(i, 0, 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, 1, n) {
scanf("%d %d", &u, &v);
E[u].pb(mp(v, i));
E[v].pb(mp(u, i));
}

dfs(1, 0);

int k;

scanf("%d", &k);
while (k--) {
scanf("%d %d", &u, &v);
fa = LCA(u, v);
++cnt[u];
++cnt[v];
cnt[fa] -= 2;
}

dfs2(1, 0);
rep(i, 1, n)
printf("%d ", ans[i]);
putchar('\n');

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

return 0;
}