Robots

\[ Time Limit: 1000 ms \quad Memory Limit: 262144 kB \]

题意

有一个机器人要从 \(1\) 点走到 \(n\) 点，每走一步都需要一天，并且这一步的代价是已经经过的天数。

思路

将问题拆成两个问题，\(dp1[i]\) 表示从 \(i\) 到 \(n\) 的天数期望，\(dp2[i]\) 表示从 \(i\) 到 \(n\) 的代价期望。
那么就很容易得到
\[ dp1[u] = \frac{\sum_{u->v}(dp1[v]) + dp1[u]}{u.size()+1} + 1\\ dp1[u] = \frac{\sum_{u->v}(dp1[v]) + u.size()+1}{u.size()} \\ dp2[u] = \frac{\sum_{u->v}(dp2[v]) + dp2[u]}{u.size()+1} + dp1[u]\\ dp2[u] = \frac{\sum_{u->v}(dp2[v]) + (u.size()+1)*dp1[u]}{u.size()} \]

/*************************************************************** 
    > File Name    : a.cpp
    > Author       : Jiaaaaaaaqi
    > Created Time : Mon 09 Sep 2019 10:07:54 PM CST
 ***************************************************************/

#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define  lowbit(x)  x & (-x)
#define  mes(a, b)  memset(a, b, sizeof a)
#define  fi         first
#define  se         second
#define  pb         push_back
#define  pii        pair<int, int>

typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 1e5 + 10;
const int    maxm = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;
using namespace std;

int n, m;
int cas, tol, T;

vector<int> vv[maxn], rv[maxn];
int ind[maxn];
double dp1[maxn], dp2[maxn];

void solve() {
    queue<int> q;
    dp1[n] = dp2[n] = 0;
    q.push(n);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        if(u != n) {
            int sz = rv[u].size();
            {
                double ans = 0;
                for(auto v : rv[u]) {
                    ans += dp1[v];
                }
                dp1[u] = 1.0*(ans+sz+1)/sz;
            }
            {
                double ans = 0;
                for(auto v : rv[u]) {
                    ans += dp2[v];
                }
                dp2[u] = 1.0*(ans+1.0*(sz+1)*dp1[u])/sz;
            }
        }
        for(auto v : vv[u]) {
            ind[v]--;
            if(ind[v] == 0) q.push(v);
        }
    }
}

int main() {
    // freopen("in", "r", stdin);
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        for(int i=1; i<=n; i++) {
            vv[i].clear();
            rv[i].clear();
            dp1[i] = dp2[i] = 0;
        }
        for(int i=1, x, y; i<=m; i++) {
            scanf("%d%d", &x, &y);
            rv[x].pb(y);
            vv[y].pb(x);
            ind[x]++;
        }
        solve();
        printf("%.2f\n", dp2[1]);
    }
    return 0;
}