一、題目描述
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 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
二、分析
增加點n為s，點n+1為t，求最大流，使用Push-Relabel算法，具體算法：最大流問題
三、代碼
1
#include<iostream>
2using namespace std;
3#define MAXN 202
4int s, t;
5int n, np, nc, m;
6char str[50];
7int c[MAXN][MAXN];
8int f[MAXN][MAXN];
9int e[MAXN];
10int h[MAXN];
11void push(int u, int v)
12
{
13
int d = min(e[u], c[u][v] - f[u][v]);
14 f[u][v] += d;
15 f[v][u] = -f[u][v];
16 e[u] -= d;
17 e[v] += d;
18
}
19bool relabel(int u)
20{
21 int mh = INT_MAX;
22 for(int i=0; i<n+2; i++)
23
{
24 if(c[u][i] > f[u][i])
25 mh = min(mh, h[i]);
26
}
27 if(mh == INT_MAX)
28 return false; //殘留網(wǎng)絡(luò)中無從u出發(fā)的路
29 h[u] = mh + 1;
30 for(int i=0; i<n+2; i++)
31 {
32 if(e[u] == 0) //已無余流，不需再次push
33 break;
34 if(h[i] == mh && c[u][i] > f[u][i]) //push的條件
35 push(u, i);
36 }
37 return true;
38}
39void init_preflow()
40{
41 memset(h, 0, sizeof(h));
42 memset(e, 0, sizeof(e));
43 h[s] = n+2;
44 for(int i=0; i<n+2; i++)
45 {
46 if(c[s][i] == 0)
47 continue;
48 f[s][i] = c[s][i];
49 f[i][s] = -f[s][i];
50 e[i] = c[s][i];
51 e[s] -= c[s][i];
52 }
53}
54void push_relabel()
55{
56 init_preflow();
57 bool flag = true; //表示是否還有relabel操作
58 while(flag)
59 {
60 flag = false;
61 for(int i=0; i<n; i++)
62 if(e[i] > 0)
63 flag = flag || relabel(i);
64 }
65}
66int main()
67{
68 while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF)
69 {
70 s = n; t = n+1;
71 memset(c, 0, sizeof(c));
72 memset(f, 0, sizeof(f));
73 while(m--)
74 {
75 scanf("%s", &str);
76 int u=0, v=0, z=0;
77 sscanf(str, "(%d,%d)%d", &u, &v, &z);
78 c[u][v] = z;
79 }
80 for(int i=0; i<np+nc; i++)
81 {
82 scanf("%s", &str);
83 int u=0, z=0;
84 sscanf(str, "(%d)%d", &u, &z);
85 if(i < np)
86 c[s][u] = z;
87 else if(i >= np && i < np + nc)
88 c[u][t] = z;
89 }
90 push_relabel();
91 printf("%d\n", e[t]);
92 }
93}
posted on 2009-06-24 19:38
Icyflame 閱讀(2098)
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