F. Flood
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100685/problem/F

Description

Input

On the first line of input integers N and K (2 ≤ N ≤ 10⁴, 0 ≤ K ≤ 10⁵) are given — the number of water ponds and the number of pond connections respectively.

On the next N lines of input integers P_i and A_i (0 ≤ A_i ≤ P_i ≤ 10⁶) are given — these are the maximum i^th water pond capacity and its initial water level.

On the next K lines of input integers F_j and T_j (1 ≤ F_j, T_j ≤ N, F_j ≠ T_j) are given — they denote a possible river flow connection from F_j to T_j water pond (reverse water flow is not possible). Consider water flow from a pond to be equally distributed between all possible flow connections from that pond. Triton is absolutely sure that there are no cycles in river flows between the ponds, and there are no multiple rivers between any two ponds.

On the last line of input integers X, Y and Z (1 ≤ X, Z ≤ N, 1 ≤ Y ≤ 10⁶) are given — the water pond that receives Triton's heavy rain, the amount of water that is added to this pond and the target pond (near the chosen town) to test respectively.

Consider that excessive water flows from a water pond if and only if its capacity is full. If some pond is overfull and no water flows are defined from that pond consider that all excessive water has flown out to the sea.

Output

The first line of the output should contain a single floating-point number L_z — the final water level in the target pond when all water flow is complete. Answers with absolute or relative error less than 10^- 4 are considered correct.

Sample Input

4 4
10 10
1 0
1 0
10 0
1 2
1 3
2 4
3 4
1 5 4

Sample Output

3.0

HINT

题意

给一个有向无环无重边的图，然后从一个地方倒水，水满了就会溢出，溢出给所有相邻的点，都平均给，然后问你最后有多少水

题解：

直接bfs就好了

代码

#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <vector>

#include <stack>

#include <map>

#include <set>

#include <queue>

#include <iomanip>

#include <string>

#include <ctime>

#include <list>

typedef unsigned char byte;

#define pb push_back

#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)

#define local freopen("in.txt","r",stdin)

#define pi acos(-1)

using namespace std;

typedef pair<double , double > dl;

const int maxn = 1e4 + ;

vector<int>e[maxn];

int n , k , z ;

int dis[maxn];

char inq[maxn];

dl A[maxn];

queue<int>q;

double ans;

double dig[maxn];

void bfs()

{

    while(!q.empty())

    {

        int cur = q.front();q.pop();

        inq[cur] = ;

        if (cur == z) return;

        int degree = e[cur].size();

        double add = (A[cur].first - A[cur].second)/((double)degree);

        A[cur].first = A[cur].second;

        for(int i =  ; i < degree ; ++ i)

        {

            int v = e[cur][i];

            A[v].first += add;

            if (A[v].first > A[v].second && !inq[v])

            {

                q.push(v);

                inq[v] = ;

            }

        }

    }

}

int main(int argc,char *argv[])

{

  scanf("%d%d",&n,&k);

  for(int i =  ; i <= n ; ++ i) scanf("%lf%lf",&A[i].second,&A[i].first);

  while(k--)

  {

      int u , v ;

      scanf("%d%d",&u,&v);

      e[u].pb(v);

  }

  int x,y;

  scanf("%d%d%d",&x,&y,&z);

  A[x].first += (double)y;

  memset(inq,,sizeof(inq));

  if (A[x].first > A[x].second) q.push(x);

  bfs();

  if (A[z].first > A[z].second ) ans = A[z].second;

  else ans = A[z].first;

  printf("%.8lf\n",ans);

  return ;

}