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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> POJ 1459 Power Network (網絡流最大流基礎 多源點多匯點 Edmonds_Karp算法)

POJ 1459 Power Network (網絡流最大流基礎 多源點多匯點 Edmonds_Karp算法)

編輯:關於C++


Power Network Time Limit: 2000MS Memory Limit: 32768K 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=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.
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An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax<喎?/kf/ware/vc/" target="_blank" class="keylink">vc3ViPih1KT15LiBUaGUgbGFiZWwgeC95IG9mIGNvbnN1bWVyIHUgc2hvd3MgdGhhdCBjKHUpPXggYW5kIGM8c3ViPm1heDwvc3ViPih1KT15LiBUaGUgbGFiZWwgeC95IG9mIHBvd2VyIHRyYW5zcG9ydCBsaW5lICh1LHYpIHNob3dzIHRoYXQgbCh1LHYpPXggYW5kIGw8c3ViPm1heDwvc3ViPih1LHYpPXkuCiBUaGUgcG93ZXIgY29uc3VtZWQgaXMgQ29uPTYuIE5vdGljZSB0aGF0IHRoZXJlIGFyZSBvdGhlciBwb3NzaWJsZSBzdGF0ZXMgb2YgdGhlIG5ldHdvcmsgYnV0IHRoZSB2YWx1ZSBvZiBDb24gY2Fubm90IGV4Y2VlZCA2Ljxicj4KCjxwIGNsYXNzPQ=="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 a 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 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.

Source

Southeastern Europe 2003

題目鏈接:poj.org/problem?id=1459

題目大意:有n個結點,np個發電站,nc個用戶,m條輸電路,給出m條輸電路及路上最大電量,然後給出np個發電站的位置及最大發電量,nc個用戶的位置及最大用電量,求電網中可被消耗的最大電量

題目分析:因為是多源點多匯點,所以我們要設置一個總源點和總匯點,使得總源點0到其他各源點,其他各匯點到總匯點n+1,然後就是普通的最大流問題,用Edmonds_Karp算法求解,把queueq寫在while(true)外面比寫在裡面快了500ms

#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));
    }
}




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