CF-goodby 2014-D

题目连接 New Year Santa Network

解法:

直接算每个边被选中的概率,

然后直接把每个边的权值*概率相加即是答案.

求每个边选中的概率可以这样求:

假设一共有n个点,那么一共有 C(n, 3)也就是 n*(n-1)*(n-2)/6种选法;

然后对于每一条边,假设他左边有x个节点,右边有y个节点,

那么选择这条边的方法有cnt = C(x, 2)*y + C(y, 2)*x种,

那选到这条边的概率就是 cnt / tot;

上面的整个过程一个DFS就搞定;

而对于每次减少某条边的权值,则直接减去权值差*概率就可以了(可以自己想想)

下面是代码

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <vector>
#include <math.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

#define FOR(i, j, k) for(int i=(j);i<=(k);i++)
#define REP(i, n) for(int i=0;i<(n);i++)
#define mst(x, y) memset(x, y, sizeof(x));
#define pii pair<int, int>
#define fr first
#define sc second
#define left myleft
#define right myright
#define ll long long
#define ull unsigned long long
#define seed 1331
#define mod ((int)1e9+7)
#define eps 1e-5
#define pdd pair<double, double>

int n;
vector <pii> edge[100009];
ll cnt[100009], len[100009], tot;
int son[100009];
double ans = 0;

ll C(ll a){
    if(a < 2) return 0;
    return a*(a-1)/2;
}

ll cal(int t){
    ll x = t, y = n - t, sum = 0;
    sum += C(x) * y;
    sum += C(y) * x;
    return sum;
}

int dfs(int u, int fa){
    son[u] = 1;
    REP(i, edge[u].size()){
        int v = edge[u][i].fr, id = edge[u][i].sc;
        if(v == fa) continue;
        int t = dfs(v, u);
        cnt[id] = cal(t);
        son[u] += t;
    }
    return son[u];
}

double D(ll x, ll y){
    double xx = x, yy = y;
    return xx / yy;
}

int main(){
   // freopen("in", "r", stdin);

    cin>>n;
    REP(i, n-1){
        int u, v, c;
        cin>>u>>v>>c;
        len[i+1] = c;
        edge[u].push_back(pii(v, i+1));
        edge[v].push_back(pii(u, i+1));
    }

    dfs(1, 0);
    tot = ((ll)n)*(n-1)*(n-2)/6;
    FOR(i, 1, n-1)
        ans += 2.0*len[i]*D(cnt[i], tot);

    int query;
    cin>>query;
    while(query -- ){
        int id, w;
        cin>>id>>w;
        ans -= 2.0*(len[id]-w)*D(cnt[id], tot);
        len[id] = w;
        printf("%.8lf\n", ans);
    }

    return 0;
}