NOIP2016模拟赛三 Problem C: 不虚就是要AK

题目大意

给定一棵带有边权的树, 问你在树上随机选两个点, 它们最短路径上的边权之和为(4)的倍数的概率为多少.

Solution

树分治. 没什么好讲的.

#include <cstdio>
#include <cctype>
#include <vector>
#include <algorithm>
#include <cstring>

using namespace std;
namespace Zeonfai
{
    inline int getInt()
    {
        int a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - '0', c = getchar();
        return a * sgn;
    }
}
const int N = (int)2e4;
int n;
int cnt[4];
long long ans;
struct tree
{
    struct edge
    {
        int v, w;
        inline edge(int _v, int _w) {v = _v; w = _w;}
    };
    struct node
    {
        vector<edge> edg;
        int vst, sz, mx;
        inline node() {vst = 0; edg.clear();}
    }nd[N + 1];
    inline void initialize() {for(int i = 1; i <= n; ++ i) nd[i] = node();}
    inline void addEdge(int u, int v, int w) {nd[u].edg.push_back(edge(v, w)); nd[v].edg.push_back(edge(u, w));}
    void getSize(int u, int pre)
    {
        nd[u].sz = 1; nd[u].mx = 0;
        for(auto edg : nd[u].edg) if(edg.v != pre && ! nd[edg.v].vst)
            getSize(edg.v, u), nd[u].sz += nd[edg.v].sz, nd[u].mx = max(nd[u].mx, nd[edg.v].sz);
    }
    int getRoot(int u, int pre, int cen)
    {
        nd[u].mx = max(nd[u].mx, nd[cen].sz - nd[u].sz);
        int res = u;
        for(auto edg : nd[u].edg) if(edg.v != pre && ! nd[edg.v].vst)
        {
            int cur = getRoot(edg.v, u, cen);
            if(nd[cur].mx < nd[res].mx) res = cur;
        }
        return res;
    }
    void getAnswer(int u, int pre, long long len)
    {
        ans += cnt[(4 - len % 4) % 4] << 1;
        for(auto edg : nd[u].edg) if(edg.v != pre && ! nd[edg.v].vst) getAnswer(edg.v, u, len + edg.w);
    }
    void update(int u, int pre, long long len)
    {
        ++ cnt[len % 4];
        for(auto edg : nd[u].edg) if(edg.v != pre && ! nd[edg.v].vst) update(edg.v, u, len + edg.w);
    }
    inline void work(int u)
    {
        getSize(u, -1);
        u = getRoot(u, -1, u);
        memset(cnt, 0, sizeof(cnt)); cnt[0] = 1; ans += 1;
        for(auto edg : nd[u].edg) if(! nd[edg.v].vst)
        {
            getAnswer(edg.v, u, edg.w);
            update(edg.v, u, edg.w);
        }
        nd[u].vst = 1;
        for(auto edg : nd[u].edg) if(! nd[edg.v].vst) work(edg.v);
    }
    inline void work() {ans = 0; work(1);}
}T;
inline void output(long long a, long long b)
{
    long long _a = a, _b = b;
    if(_a < _b) swap(_a, _b);
    while(_b)
    {
        long long tmp = _b;
        _b = _a % _b;
        _a = tmp;
    }
    printf("%lld/%lld
", a / _a, b / _a);
}
int main()
{

#ifndef ONLINE_JUDGE

    freopen("AK.in", "r", stdin);
    freopen("AK.out", "w", stdout);

#endif

    using namespace Zeonfai;
    while(n = getInt())
    {
        T.initialize();
        for(int i = 1, u, v, c; i < n; ++ i) u = getInt(), v = getInt(), c = getInt(), T.addEdge(u, v, c);
        T.work();
        output(ans, (long long)n * n);
    }
}