Power Network
| Time Limit: 2000MS | Memory Limit: 32768K | |
| Total Submissions: 22517 | Accepted: 11800 |
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.
An example is 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 are other possible states of the network but the value of Con cannot exceed 6.
An example is 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 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.
第一道网络流的题。必须加两个点作为源点和汇点。（a，b）c 相当与a--b间的容量为c
（a）b相当于从源点到a点的容量或者从a点到汇点的容量，建图后，用dinic即可求出。
1 /*====================================================================== 2 * Author : kevin 3 * Filename : PowerNetwork.cpp 4 * Creat time : 2014-07-16 09:41 5 * Description : 6 ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio> 10 #include <cstring> 11 #include <queue> 12 #include <cmath> 13 #define clr(a,b) memset(a,b,sizeof(a)) 14 #define M 105 15 #define INF 0x7f7f7f7f 16 using namespace std; 17 struct Edge{ 18 int from,to,cap,flow; 19 }; 20 int n,np,nc,m,supers,supert; 21 vector<Edge> edges; 22 vector<int> G[M]; 23 bool vis[M]; 24 int d[M],cur[M]; 25 bool BFS() 26 { 27 clr(vis,0); 28 queue<int>Q; 29 Q.push(supers); 30 d[supers] = 0; 31 vis[supers] = 1; 32 while(!Q.empty()){ 33 int x = Q.front(); Q.pop(); 34 int len = G[x].size(); 35 for(int i = 0; i < len; i++){ 36 Edge& e = edges[G[x][i]]; 37 if(!vis[e.to] && e.cap > e.flow){ 38 vis[e.to] = 1; 39 d[e.to] = d[x]+1; 40 Q.push(e.to); 41 } 42 } 43 } 44 return vis[supert]; 45 } 46 int DFS(int x,int a) 47 { 48 if(x == supert || a == 0) return a; 49 int flow = 0,f; 50 int len = G[x].size(); 51 for(int& i = cur[x]; i < len; i++){ 52 Edge& e = edges[G[x][i]]; 53 if(d[x] + 1 == d[e.to] && (f = DFS(e.to,min(a,e.cap-e.flow))) > 0){ 54 e.flow += f; 55 edges[G[x][i]^1].flow -= f; 56 flow += f; 57 a -= f; 58 if(a == 0) break; 59 } 60 } 61 return flow; 62 } 63 void AddEdge(int from,int to,int cap) 64 { 65 edges.push_back((Edge) {from,to,cap,0}); 66 edges.push_back((Edge) {to,from,0,0}); 67 int m = edges.size(); 68 G[from].push_back(m-2); 69 G[to].push_back(m-1); 70 } 71 int Dinic(int s,int t) 72 { 73 int flow = 0; 74 while(BFS()){ 75 clr(cur,0); 76 flow += DFS(s,INF); 77 } 78 return flow; 79 } 80 int main(int argc,char *argv[]) 81 { 82 int u,v,cap; 83 while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF){ 84 char str[10]; 85 for(int i = 0; i < m; i++){ 86 scanf("%s",str); 87 sscanf(str,"(%d,%d)%d",&u,&v,&cap); 88 AddEdge(u,v,cap); 89 } 90 supers = n; 91 int num; 92 for(int i = 0; i < np; i++){ 93 scanf("%s",str); 94 sscanf(str,"(%d)%d",&num,&cap); 95 AddEdge(supers,num,cap); 96 } 97 supert = n+1; 98 for(int i = 0; i < nc; i++){ 99 scanf("%s",str); 100 sscanf(str,"(%d)%d",&num,&cap); 101 AddEdge(num,supert,cap); 102 } 103 int ans = Dinic(supers,supert); 104 printf("%d\n",ans); 105 edges.clear(); 106 for(int i = 0; i < M; i++) 107 G[i].clear(); 108 } 109 return 0; 110 }