Power Network
| Time Limit: 2000MS |
|
Memory Limit: 32768K |
| Total Submissions: 16422 |
|
Accepted: 8712 |
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.
練題目/最大流增廣路（km）/1459%20--%20Power%20Network.files/1459_1.jpg)
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
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.
哎，糾結(jié)死了，我對(duì)網(wǎng)絡(luò)流這方面理解的還不行
如果自己寫代碼的話還是有點(diǎn)難度，所以找個(gè)好的模版還是很重要的
額，模版也比較糾結(jié)，好多中算法
找了個(gè)比較簡(jiǎn)單的算法 Edmonds_karp
時(shí)間復(fù)雜度為O(V*E^2)
Edmonds-Karp算法就是利用寬度優(yōu)先不斷地找一條從s到t的可改進(jìn)路，然后改進(jìn)流量，一直到找不到可改進(jìn)路為止。
由于用寬度優(yōu)先，每次找到的可改進(jìn)路是最短的可改進(jìn)路，通過(guò)分析可以知道其復(fù)雜度為O(VE^2)。
代碼好丑
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define MAX 105
5int map[MAX][MAX],flow[MAX][MAX],c[MAX][MAX];
6int n,nc,np,nt,s,t;
7int sum;
8int min(int a,int b)
9
{
10
if (a<b) return a;else return b;
11
}
12void Edmonds_Karp()
13{
14 int l1[MAX],l2[MAX],q[MAX];
15 int u,v,head,tail;
16 do
17
{
18 memset(l1,0,sizeof(l1));
19 memset(l2,0,sizeof(l2));//初始化所有標(biāo)號(hào)為0
20 l1[s]=0;l2[s]=0x7fffffff;
21 head=0;tail=1;
22 q[tail]=s;
23 while (head<tail&&l2[t]==0)//q未空且匯點(diǎn)未標(biāo)號(hào)
24 {
25 head++;
26 u=q[head];
27 for (v=1;v<=n ;v++ )
28 {
29 if (flow[u][v]<c[u][v]&&l2[v]==0)//未標(biāo)號(hào)且有可行流
30 {
31 tail++;
32 q[tail]=v;
33 l2[v]=min(c[u][v]-flow[u][v],l2[u]);
34 //l2[v]記錄s到v增廣路中最小的可改進(jìn)流
35 l1[v]=u;//記錄前驅(qū)
36
}
37 }
38 }
39 if (l2[t]>0)//匯點(diǎn)未標(biāo)號(hào)
40 {
41 v=t;
42 u=l1[v];
43 while (v!=s)
44 {
45 flow[u][v]+=l2[t];
46 flow[v][u]=-flow[u][v];
47 v=u;
48 u=l1[v];
49 }
50 }
51 }
52 while (l2[t]!=0);//直到匯點(diǎn)未標(biāo)號(hào)
53}
54void init()
55{
56 int i,j,a,b,w,x;
57 char ch1;
58 s=1;t=n+2;
59 memset(map,0,sizeof(map));
60 for (i=1;i<=nt ;i++ )
61 {
62 scanf("%c",&ch1);
63 while (ch1!='(')
64 {
65 scanf("%c",&ch1);
66 }
67 scanf("%d",&a);a=a+2;
68 scanf("%c",&ch1);scanf("%d",&b);b=b+2;
69 scanf("%c",&ch1);scanf("%d",&w);
70 map[a][b]=w;
71 }
72 for (i=1;i<=np ;i++ )
73 {
74 scanf("%c",&ch1);
75 while (ch1!='(')
76 {
77 scanf("%c",&ch1);
78 }
79 scanf("%d",&a);a=a+2;
80 scanf("%c",&ch1);scanf("%d",&w);
81 map[s][a]=w;//map[a][s]=-w;
82 }
83 for (i=1;i<=nc ;i++ )
84 {
85 scanf("%c",&ch1);
86 while (ch1!='(')
87 {
88 scanf("%c",&ch1);
89 }
90 scanf("%d",&a);a=a+2;
91 scanf("%c",&ch1);scanf("%d",&w);
92 map[a][t]=w;//map[t][a]=-w;
93 }
94 n=n+2;
95 /**//*/for (i=1;i<=n ;i++ )
96 {
97 for (j=1;j<=n;j++ )
98 {
99 printf("%d ",map[i][j]);
100 }
101 printf("\n");
102 }*/
103 for (i=1;i<=n ;i++ )
104 {
105 for (j=1;j<=n;j++ )
106 {
107 c[i][j]=map[i][j];
108 }
109 }
110}
111int main()
112{
113 int i;
114 while (scanf("%d%d%d%d",&n,&np,&nc,&nt)!=EOF)
115 {
116 memset(flow,0,sizeof(flow));
117 init();
118 Edmonds_Karp(1,n);
119 sum=0;
120 for (i=1;i<=n;i++ )
121 {
122 sum+=flow[1][i];
123 }
124 printf("%d\n",sum);
125 }
126 return 0;
127}
128