POJ 1459 Power Network (the Edmonds_Karp Algorithm for Multi-source and multi-sink points of the maximum flow of Network streams)

POJ 1459 Power Network (the Edmonds_Karp Algorithm for Multi-source and multi-sink points of the maximum flow of Network streams)

Power Network

Time Limit:2000 MS		Memory Limit:32768 K
Total Submissions:24056		Accepted:12564

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s (u)> = 0 of power, may produce an amount 0 <= p (u) <= pmax (u) of power, may consume an amount 0 <= c (u) <= min (s (u), cmax (u) of power, and may deliver an amount d (u) = s (u) + p (u)-c (u) of power. the following restrictions apply: c (u) = 0 for any power station, p (u) = 0 for any consumer, and p (u) = c (u) = 0 for any dispatcher. there is at most one power transport line (u, v) from a node u to a node v in the net; it transports an amount 0 <= l (u, v) <= lmax (u, v) of power delivered by u to v. let Con = Σ uc (u) be the power consumed in the net. the problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p (u) = x and pmax <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> Vertex = "pst"> Input

There are several data sets in the input. each data set encodes a power network. it starts with four integers: 0 <=n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers ), and 0 <= m <= n ^ 2 (power transport lines ). follow m data triplets (u, v) z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax (u, v ). follow np doublets (u) z, where u is the identifier of the power station and 0 <= z <= 10000 is the value of pmax (u ). the data set ends with nc doublets (u) z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax (u ). all input numbers are integers. except t the (u, v) z triplets and the (u) z doublets, which do not contain white spaces, white spaces can occur freely in input. input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)207 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7         (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5         (0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

156

Hint

The sample input contains two data sets. the first data set encodes a network with 2 nodes, power station 0 with pmax (0) = 15 and consumer 1 with cmax (1) = 20, and 2 power transport lines with lmax (0, 1) = 20 and lmax (1, 0) = 10. the maximum value of Con is 15. the second data set encodes the network from figure 1.

Source

Southeastern Europe 2003

Link: poj.org/problem? Id = 1459

There are n nodes, np power stations, nc users, m transmission routes, m transmission routes, and the maximum power consumption on the routes, then, the location and maximum power generation of the np power station are given. The location of the nc user and the maximum power consumption are obtained to find the maximum power consumption in the power grid.

Question Analysis: because it is a multi-source Multi-sink point, we need to set a total source point and a total sink point, so that the total source point 0 to other source points, other sink points to the total sink point n + 1, then, we use the Edmonds_Karp algorithm to solve the common largest stream problem. Q writing outside while (true) is 500 ms faster than writing in it

#include 
   
    #include 
    
     #include 
     
      #include using namespace std;int const INF = 0x3fffffff;int const MAX = 105;int c[MAX][MAX];int f[MAX][MAX];int a[MAX];int pre[MAX];int n, np, nc, m;int Edmonds_Karp(int s, int t){    int ans = 0;    queue 
      
        q;    while(true)    {        memset(a, 0, sizeof(a));        a[s] = INF;        q.push(s);        while(!q.empty())        {            int u = q.front();            q.pop();            for(int v = 0; v <= n + 1; v++)            {                if(!a[v] && c[u][v] > f[u][v])                {                    a[v] = min(a[u], c[u][v] - f[u][v]);                    pre[v] =  u;                    q.push(v);                }            }        }        if(a[t] == 0)            break;        for(int u = t; u != s; u = pre[u])        {            f[pre[u]][u] += a[t];            f[u][pre[u]] -= a[t];        }        ans += a[t];    }    return ans;}int main(){    while(scanf("%d %d %d %d", &n, &np, &nc, &m) != EOF)    {        memset(c, 0, sizeof(c));        memset(f, 0, sizeof(f));        for(int i = 0 ; i < m; i++)        {            int u, v, w;            scanf(" (%d,%d)%d", &u, &v, &w);            c[u + 1][v + 1] += w;        }        for(int i = 0; i < np; i++)        {            int v, w;            scanf(" (%d)%d", &v, &w);            c[0][v + 1] += w;        }        for(int i = 0; i < nc; i++)        {            int u, w;            scanf(" (%d)%d", &u, &w);            c[u + 1][n + 1] += w;        }        printf("%d\n", Edmonds_Karp(0, n + 1));    }}