Gym

 Gym - 100685F

题意：n个水池之间流水，溢出多少流出多少，多个流出通道的话平均分配，给你每个水池中的水量和容量，问到最后目标水池中水量。

思路：直接用队列扩展，不过这里有一个优化，就是统计一下每个点的入度，只有对一个点访问次数达到入度次了，再将其加入队尾，这样就保证了对每个点只操作一次，不然WA、TLE各种错

#pragma comment(linker, "/STACK:1000000000")
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define LL long long
#define INF 0x3f3f3f3f
#define MAXN 100005
#define MAXK 100005
#define eps 1e-8
using namespace std;
struct Node{
    double m, now;
};
Node p[MAXK];
int cnt[MAXK];
vector<int> G[MAXK];
queue<int> Q;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // OPEN_FILE
    int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i++){
        scanf("%lf%lf", &p[i].m, &p[i].now);
        //p[i].pos = i;
    }
    int x, y, z;
    //double y;
    memset(cnt, 0, sizeof(cnt));
    for(int i = 1; i <= k; i++){
        scanf("%d%d", &x, &z);
        G[x].push_back(z);
        cnt[z]++;
    }
    scanf("%d%d%d", &x, &y, &z);
   // int head = 1, tail = 1;
    p[x].now += y;
    Q.push(x);
    //vis[x] = true;
    //double ans = p[z].now;
    //int cnt = 0;
    while(!Q.empty()){
        //cnt++;
        int pos = Q.front();
        Q.pop();
        Node q = p[pos];
        if(q.now <= q.m) continue;
        p[pos].now = p[pos].m;
        if(G[pos].size() == 0) continue;
        double u =  (q.now - q.m) / G[pos].size();
        for(int i = 0; i < G[pos].size(); i++){
            //if(vis[G[q.pos][i]]) continue;
            p[G[pos][i]].now += u;
            //tail++;
            cnt[G[pos][i]]--;
            if(cnt[G[pos][i]] == 0){
                Q.push(G[pos][i]);
            }
        }
    }
    printf("%.6lf
", p[z].now);
}