【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。

  1 /* 191C */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23 
 24 #define sti                set<int>
 25 #define stpii            set<pair<int, int> >
 26 #define mpii            map<int,int>
 27 #define vi                vector<int>
 28 #define pii                pair<int,int>
 29 #define vpii            vector<pair<int,int> >
 30 #define rep(i, a, n)     for (int i=a;i<n;++i)
 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 32 #define clr                clear
 33 #define pb                 push_back
 34 #define mp                 make_pair
 35 #define fir                first
 36 #define sec                second
 37 #define all(x)             (x).begin(),(x).end()
 38 #define SZ(x)             ((int)(x).size())
 39 #define lson            l, mid, rt<<1
 40 #define rson            mid+1, r, rt<<1|1
 41 
 42 const int maxn = 1e5+5;
 43 const int maxd = 17;
 44 vpii E[maxn];
 45 bool visit[maxn];
 46 int ans[maxn];
 47 int cnt[maxn];
 48 int deep[maxn];
 49 int  pre[maxn][maxd];
 50 
 51 void dfs(int u, int fa) {
 52     int sz = SZ(E[u]), v;
 53     
 54     deep[u] = deep[fa] + 1;
 55     rep(i, 0, sz) {
 56         v = E[u][i].fir;
 57         if (v == fa)
 58             continue;
 59         pre[v][0] = u;
 60         rep(j, 1, maxd)
 61             pre[v][j] = pre[pre[v][j-1]][j-1];
 62         dfs(v, u);
 63     }
 64 }
 65 
 66 int LCA(int u, int v)  {
 67     if (deep[u] > deep[v])
 68         swap(u, v);
 69     if (deep[u] < deep[v]) {
 70         int tmp = deep[v] - deep[u];
 71         rep(i, 0, maxd) {
 72             if (tmp & (1<<i))
 73                 v = pre[v][i];
 74         }
 75     }
 76     
 77     if (u != v) {
 78         per(i, 0, maxd) {
 79             if (pre[u][i] != pre[v][i]) {
 80                 u = pre[u][i];
 81                 v = pre[v][i];
 82             }
 83         }
 84         return pre[u][0];
 85     } else {
 86         return u;
 87     }
 88 }
 89 
 90 int dfs2(int u, int eid) {
 91     int ret = 0, sz = SZ(E[u]);
 92     int e, v;
 93     
 94     visit[u] = true;
 95     rep(i, 0, sz) {
 96         v = E[u][i].fir;
 97         e = E[u][i].sec;
 98         if (!visit[v]) {
 99             ret += dfs2(v, e);
100         }
101     }
102     ret += cnt[u];
103     ans[eid] = ret;
104     
105     return ret;
106 }
107 
108 int main() {
109     ios::sync_with_stdio(false);
110     #ifndef ONLINE_JUDGE
111         freopen("data.in", "r", stdin);
112         freopen("data.out", "w", stdout);
113     #endif
114     
115     int n;
116     int u, v, fa;
117     
118     scanf("%d", &n);
119     rep(i, 1, n) {
120         scanf("%d %d", &u, &v);
121         E[u].pb(mp(v, i));
122         E[v].pb(mp(u, i));
123     }
124     
125     dfs(1, 0);
126     
127     int k;
128     
129     scanf("%d", &k);
130     while (k--) {
131         scanf("%d %d", &u, &v);
132         fa = LCA(u, v);
133         ++cnt[u];
134         ++cnt[v];
135         cnt[fa] -= 2;
136     }
137     
138     dfs2(1, 0);
139     rep(i, 1, n)
140         printf("%d ", ans[i]);
141     putchar('\n');
142     
143     #ifndef ONLINE_JUDGE
144         printf("time = %d.\n", (int)clock());
145     #endif
146     
147     return 0;
148 }

 

posted on 2015-09-23 22:11  Bombe  阅读(266)  评论(0)  编辑  收藏  举报

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