poj_1459 Power Network(多源多匯最大流)

 

 

 

1459Accepted676K641MSG++1846B2014-07-16 22:22:30

Power Network

Time Limit: 2000MS Memory Limit: 32768KTotal Submissions: 22488 Accepted: 11778

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 l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(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 c_max(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)207 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

156

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

 

題目大意，給出從0開始編號的n個點，其中有發電站，中轉站，用戶，發電站和用戶都有容量上限，要求在這個電網中所能傳輸的最大電量。

網絡流的題目，一開始input的時候沒想到怎麼讀取，因為題目說除了給出的有括號的容量信息沒空格時，其他空格都是任意的，一開始用scanf("(%d,%d)%d",&u,&v,&w);來讀取，但是可能是讀到了空格，出現了死循環。然後看了題解的讀取方法，先用while循環消除空格，讀到第一個左括號，再用scanf()的方法來讀接下來的。

這題是多源點多匯點的，可以加一個超級源點和超級匯點，超級源點和之前源點的邊容量為之前源點的容量，之前匯點與超級匯點的邊容量亦然。

 

#include<cstdio>#include<cstdlib>#include<cstring>#include<queue>using namespace std;#define range 110#define INF 1000000  int flow[range][range],cap[range][range],p[range],a[range];int E_K(int s,int t)     //做最大流{   queue<int>q;   int i;   int ans=0;   while(!q.empty()) q.pop(); while(1)  {   memset(a,0,sizeof(a));   a[s]=INF;//   printf("en\n");   q.push(s);   while(!q.empty())   {      int u=q.front(); q.pop();      for(i=0;i<=t;i++)      {         if(!a[i] && cap[u][i]-flow[u][i]>0)         {            p[i]=u;            q.push(i);            a[i]=cap[u][i]-flow[u][i];             if(a[u]<a[i]) a[i]=a[u];         }      }   }      if(a[t]==0) break;      for(i=t;i!=s;i=p[i])      {         flow[p[i]][i]+=a[t];         flow[i][p[i]]-=a[t];   //更新逆向流      }      ans+=a[t];  }   return ans;}int main(){   freopen("in.txt","r",stdin);   int n,np,nc,m,i,j;   int ms[range],mt[range];   while(scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF)   {//      printf("n=%d np=%d nc=%d m=%d\n",n,np,nc,m);      memset(flow,0,sizeof(flow));      memset(cap,0,sizeof(cap));      int u,v,w;      char c;      int s,t;      s=n; t=n+1;      for(i=0;i<m;i++)      {         while(getchar()!='(');     //input做法，先用while循環將空格消除掉         scanf("%d,%d)%d",&u,&v,&w);         cap[u][v]=w;//         printf("%d %d %d\n",u,v,w);      }      for(i=0;i<np;i++)      {         while(getchar()!='(');         scanf("%d)%d",&u,&v);         cap[s][u]=v; //        ms[u]=v; //        printf("%d %d\n",u,v);      }      for(i=0;i<nc;i++)      {         while(getchar()!='(');         scanf("%d)%d",&u,&v);         cap[u][t]=v;//         mt[u]=v;//         printf("%d %d\n",u,v);      }      int ans=E_K(s,t);      printf("%d\n",ans);   }   return 0;}