Gym - 100685F Flood BFS

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 = ; i <= n; i++){
         scanf("%lf%lf", &p[i].m, &p[i].now);
         //p[i].pos = i;
     }
     int x, y, z;
     //double y;
     memset(cnt, , sizeof(cnt));
     for(int i = ; 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() == ) continue;
         double u =  (q.now - q.m) / G[pos].size();
         for(int i = ; 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]] == ){
                 Q.push(G[pos][i]);
             }
         }
     }
     printf("%.6lf\n", p[z].now);
 }