Power Network
| Time Limit: 2000MS | Memory Limit: 32768K | |
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.
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.
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.
(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.
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
题意：有多个电站和多个消费者，求消费者消费的最大电能。
可以引入一个超级源点s和一个超级汇点t，源点和每一个电站建一条边，边容量为电站的最大产电量；每个消费者和汇点建一条边，边容量为消费者消费的最大电能。
#include <iostream> #include<cstring> #include<cstdio> using namespace std; const int MAXN=105; const int INF=(1<<29); int level[MAXN]; int flow[MAXN][MAXN]; int s,t,n,np,nc,m; bool bfs() { int queue[MAXN],front,rear; front=rear=0; memset(level,0,sizeof(level)); level[s]=1; queue[rear++]=s; while(front!=rear) { int v=queue[front++]; for(int i=0;i<=n;i++) if(!level[i]&&flow[v][i]) { level[i]=level[v]+1; queue[rear++]=i; } } if(!level[t]) return false; return true; } int dfs(int i,int f) { if(i==t) return f; int sum=0; for(int j=0;f&&j<=n;j++) { if(flow[i][j]&&level[i]+1==level[j]) { int tmp=dfs(j,min(f,flow[i][j])); flow[i][j]-=tmp; flow[j][i]+=tmp; sum+=tmp; f-=tmp; } } return sum; } int dinic() { int maxflow=0; while(bfs()) maxflow+=dfs(s,INF); return maxflow; } void build_graph() { int u,v,f; char str[10]; memset(flow,0,sizeof(flow)); while(m--) { scanf("%s",str); sscanf(str,"(%d,%d)%d",&u,&v,&f); flow[u][v]=f; } while(np--) { scanf("%s",str); sscanf(str,"(%d)%d",&u,&f); flow[s][u]=f; } while(nc--) { scanf("%s",str); sscanf(str,"(%d)%d",&u,&f); flow[u][t]=f; } } int main() { while(~scanf("%d%d%d%d",&n,&np,&nc,&m)) { n++; s=n-1; t=n; build_graph(); printf("%d\n",dinic()); } return 0; } /* 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 */