POJ 1459 Power Network

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) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(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) <= l_max(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 p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6. 

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 l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(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 c_max(u). All input numbers are integers. Except 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)20
7 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:

15
6

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.

 

题意：有m条电线，np个发电厂，nc个耗电厂，（u，v）z表示从u点能传到v点的电量是z，np个（u）z表示u点最多发电z，nc个（u）z表示u点最多消耗电量z，问消耗的最大电量是多少。

 

注意输入

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;

const int INF=0x3f3f3f3f;
const int N=110;

int G[N][N], layer[N], a, b, c, A, B;

void Change(char s[])
{
    int i;

    for (i = 0; s[i] != ''; i++)
    {
        if (s[i] == ',') A = i;
        if (s[i] == ')') B = i;
    }

    for (i = 1; i < A; i++) a = a*10 + (s[i]-'0');
    for (i = A+1; i < B; i++) b = b*10 + (s[i]-'0');
    for (i = B+1; s[i] != ''; i++) c = c*10 + (s[i]-'0');
}
int BFS(int Start, int End)
{
    int i, u;
    queue<int>Q;

    memset(layer, 0, sizeof(layer));

    Q.push(Start);
    layer[Start] = 1;

    while (!Q.empty())
    {
        u = Q.front(); Q.pop();

        if (u == End) return 1;

        for (i = 0; i <= End; i++)
        {
            if (G[u][i] && !layer[i])
            {
                layer[i] = layer[u]+1;
                Q.push(i);
            }
        }
    }

    return 0;
}

int DFS(int u, int End, int mini)
{
    int i, ans, con = 0;

    if (u == End) return mini;

    for (i = 0; i <= End; i++)
    {
        if (G[u][i] && layer[i] == layer[u]+1)
        {
            ans = min(G[u][i], mini-con);
            ans = DFS(i, End, ans);

            G[u][i] -= ans;
            G[i][u] += ans;
            con += ans;

            if (con == mini) break;
        }
    }

    if (con == 0) layer[u] = 0;

    return con;
}

int Dinic(int Start, int End)
{
    int ans = 0;

    while (BFS(Start, End))
        ans += DFS(Start, End, INF);

    return ans;
}
int main ()
{
    int n, np, nc, m, S, E, ans;
    char s[30];

    while (scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)
    {
        S = n; E = n+1;
        memset(G, 0, sizeof(G));

        while (m--)
        {
            a = b = c = A = B = 0;
            scanf("%s", s);
            Change(s);
            G[a][b] += c;
        }
        while (np--)
        {
            a = b = c = A = B = 0;
            scanf("%s", s);
            Change(s);
            G[S][b] += c;
        }
        while (nc--)
        {
            a = b = c = A = B = 0;
            scanf("%s", s);
            Change(s);
            G[b][E] += c;
        }

        ans = Dinic(S, E);

        printf("%d
", ans);
    }

    return 0;
}