CF-goodby 2014-D

来源：互联网发布：八宝茶的软件编辑：程序博客网时间：2024/05/01 18:19

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;}

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