luogu P2634 [国家集训队]聪聪可可

传送门

上来想的树形dp怎么破......

因为点分治做到的就是把树上所有路径问题转化成过一点的路径问题

所以可以分开统计过某一点的路径

记录%3之后的路径条数就可以做 分析一下就出来了

这里注意向下递归的时候要把方案数减掉 也就是74行那个ans-calc()

剩下的套点分治板子就星

Code:

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define ms(a,b) memset(a,b,sizeof a)
 5 #define rep(i,a,n) for(int i = a;i <= n;i++)
 6 #define per(i,n,a) for(int i = n;i >= a;i--)
 7 #define inf 2147483647
 8 using namespace std;
 9 typedef long long ll;
10 ll read() {
11     ll as = 0,fu = 1;
12     char c = getchar();
13     while(c < '0' || c > '9') {
14         if(c == '-') fu = -1;
15         c = getchar();
16     }
17     while(c >= '0' && c <= '9') {
18         as = as * 10 + c - '0';
19         c = getchar();
20     }
21     return as * fu;
22 }
23 //head
24 const int N = 100003;
25 int n,m,q[N];
26 int head[N],mo[N<<1],nxt[N<<1],cst[N<<1],cnt;
27 void addd(int x,int y,int w) {
28     nxt[++cnt] = head[x],head[x] = cnt;
29     mo[cnt] = y,cst[cnt] = w;
30 }
31 void add(int x,int y) {
32     int w = read() % 3;
33     if(x^y) addd(x,y,w),addd(y,x,w);
34 }
35 
36 int sze[N];
37 bool vis[N];
38 int sum,rt,maxx[N];
39 void getroot(int x,int f) {
40     sze[x] = 1,maxx[x] = 0;
41     for(int i = head[x];i;i = nxt[i]) {
42         int sn = mo[i];
43         if(vis[sn] || sn == f) continue;
44         getroot(sn,x),sze[x] += sze[sn];
45         maxx[x] = max(maxx[x],sze[sn]);
46     }
47     maxx[x] = max(maxx[x],sum - sze[x]);
48     if(maxx[x] < maxx[rt]) rt = x;
49 }
50 
51 int dis[N];
52 int con[3];
53 void getdis(int x,int f) {
54     con[dis[x]]++;
55     for(int i = head[x];i;i = nxt[i]) {
56         int sn = mo[i];
57         if(sn == f || vis[sn]) continue;
58         dis[sn] = (dis[x] + cst[i]) % 3;
59         getdis(sn,x);
60     }
61 }
62 
63 int ans;
64 int calc(int x,int d) {
65     dis[x] = d,ms(con,0),getdis(x,x);
66     return con[0] * con[0] + con[1] * con[2] * 2;
67 }
68 
69 void solve(int x) {
70     vis[x] = 1,ans += calc(x,0);
71     for(int i = head[x];i;i = nxt[i]) {
72         int sn = mo[i];
73         if(vis[sn]) continue;
74         ans -= calc(sn,cst[i]);
75         sum = sze[sn],maxx[rt = 0] = inf;
76         getroot(sn,sn),solve(rt);
77     }
78 }
79 
80 int gcd(int x,int y) {return y ? gcd(y,x%y) : x;}
81 
82 int main() {
83     n = read();
84     rep(i,2,n) add(read(),read());
85     sum = n,maxx[rt = 0] = inf;
86     getroot(1,1),solve(rt);
87     int d = gcd(ans,n*n);
88     printf("%d/%d\n",ans/d,n*n/d);
89     return 0;
90 }

 

posted @ 2018-12-01 15:23  白怀潇  阅读(125)  评论(0)  编辑  收藏  举报