POJ 1459 Power Network Classic stream composition problem Max stream, EK algorithm

Title Link: POJ 1459 Power Network

Power Network Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 23347 Accepted: 12231 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node U is supplied with an amount s (U) >= 0 of the power, may produce an amount 0 <= P (u) <= pmax (U) of power, May consume a 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 Powe R. 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 Dispat Cher. 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 the Power delivered by U v. Let Con=σuc (U) is the power consumed in the net. The problem is to compute the maximum value of Con. An example was in Figure 1. The label X/y of power station U shows that P (u) =x and Pmax (U) =y. The label x/y of consumer U shows that C (U) =x and Cmax (U) =y. The label X/y of Power Transport Line (U,V) shows that L (u,v) =x and Lmax (u,v) =y. The power consumed is con=6. Notice that there is other possible states of the network but the value of Con cannot exceed 6. Input There is several data sets in the input. Each data set encodes a power network. It starts with four integers:0 <= n <= (nodes), 0 <= NP <= N (power stations), 0 <= NC <= N (consum ERS), 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 <= of Lmax (U,V). Follow NP doublets (U) z, where U is the identifier of a power station and 0 <= Z <= 10000 is the value of Pmax (U). The data set ends with NC doublets (u) z, where u are the identifier of a consumer and 0 <= Z <= 10000 is the value of Cmax (U). All input numbers is integers. Except the (u,v) z Triplets and the (U) z doublets, which do not contain white spaces, white spaces can occur freely in Inpu T. Input data terminate with an end of file and is correct. Output For each data set from the input, the program prints on the standard output the maximum amount of power that can is consum Ed in the corresponding network. Each result have an integral value and are printed from the beginning of a separate line. Sample Input 2 1 1 2 (0,1) (1,0) ten (0) (1) 207 2 3 (0,0) 1 (0,1) 2 (0,2) 5 (1,0) 1 (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 the 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 p Ower 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

Test instructions

A grid consists of n nodes, which are connected by wires.

There are three types of nodes:

(1) Power plant: generating electricity and transmitting electricity

(2) Consumers: Energy consumption, can also transmit electricity

(3) Dispatch station: does not produce electricity, does not consume the electricity, transmits the electricity only.

The maximum consumption of electricity is now demanded. There are multiple power plants and consumers.

Analysis:

Add a new point as the source point, an arc to each power station, and a weight of maximum production capacity

A new point is added as a meeting point, and each consumer gives it an arc, which is the maximum power consumption.

All other points, no matter what, can be regarded as the dispatch station, that is, enter equals output.

The composition is completed, the maximum flow algorithm is used to solve the problem, take up the EK algorithm, a high degree of complexity.

Code:

#include <iostream> #include <cstdio> #include <cstring> #include <queue>using namespace std;#  Define MAXN 110#define INF 0x3f3f3f3fint ans, s, T, N;int A[MAXN], pre[maxn];int flow[maxn][maxn];int cap[maxn][maxn];void    Edmonds_karp () {queue<int> q;    memset (flow, 0, sizeof (flow));    Ans = 0;        while (1) {memset (A, 0, sizeof (a));        A[s] = INF;        Q.push (s);            while (!q.empty ())//bfs find the augmented path {int u = q.front ();            Q.pop ();                    for (int v = 0; v < n; v++) if (!a[v] && cap[u][v] > Flow[u][v]) {                    PRE[V] = u;                    Q.push (v);                A[v] = min (A[u], cap[u][v]-flow[u][v]);        }} if (a[t] = = 0) break;            for (int u = t; u! = s; u = Pre[u])//Improved network stream {Flow[pre[u]][u] + = a[t];        Flow[u][pre[u]]-= a[t];    } ans + = a[t]; }}int Main () {//freopen ("poj_1459. txt "," R ", stdin);    int m, u, V, C, NP, NC;        while (~SCANF ("%d%d%d%d", &n, &AMP;NP, &AMP;NC, &m)) {s = n, t = n+1;        n + = 2;        cout << s << "<< t <<" "<< n << Endl;        memset (Cap, 0, sizeof (CAP));            while (m--) {while (GetChar ()! = ' (');            scanf ("%d,%d)%d", &u, &v, &c);        CAP[U][V] = c;            } while (np--) {while (GetChar ()! = ' (');            scanf ("%d)%d", &v, &c);        CAP[S][V] = c;            } while (nc--) {while (GetChar ()! = ' (');            scanf ("%d)%d", &u, &c);        Cap[u][t] = c; }/* for (int i = 0, i < n; i++) {for (int j = 0; J < N; j + +) cout &lt            ;< Cap[i][j] << "";        cout << Endl;        } */Edmonds_karp ();    printf ("%d\n", ans); } return 0;}

POJ 1459 Power Network Classic stream composition problem Max stream, EK algorithm