POJ 1459 Power Network(网络流)

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.

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)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

题目大意

一共有n个站点，其中np个发电站，nc个消耗站，剩余的为中转站。发电站只发电，消耗战只消耗电，中转站只是传送电，然后给m根电线及其传电能力，问所有的消耗站能获得的最大的电能是多少。

解题思路

该题中有多个源点和汇点，所以增加一个超级源点和一个超级汇点，将所有的发电站都与超级源点连接起来，所有的消耗站都与超级汇点连接起来，便转化为了单源单汇模型。用EK求最大流即可。

代码实现

#include <iostream>#include<queue>#include<cstdio>#include<cstring>using namespace std;#define maxn 107*2#define maxx 100000int maps[maxn][maxn],pre[maxn];int n,np,nc,m;bool EK_bfs(int s,int e){    queue<int>qu;    bool flag[maxn];    memset(flag,false,sizeof(flag));    memset(pre,-1,sizeof(pre));    qu.push(s);    flag[s]=1;    while(!qu.empty())    {        int t=qu.front();        qu.pop();        if(t==e)  return true;        for(int i=1; i<=n+1; i++)        {            if(maps[t][i]&&!flag[i])            {                flag[i]=1;                pre[i]=t;                qu.push(i);            }        }    }    return false;}int EK_Max_flow(int s,int e){    int u,flow_ans=0,minn;    while(EK_bfs(s,e))    {        minn=maxx;        u=e;        while(pre[u]!=-1)        {            minn=min(minn,maps[pre[u]][u]);            u=pre[u];        }        flow_ans+=minn;        u=e;        while(pre[u]!=-1)        {            maps[pre[u]][u]-=minn;            maps[u][pre[u]]+=minn;            u=pre[u];        }    }    return flow_ans;}int main(){    while(~scanf("%d %d %d %d",&n,&np,&nc,&m))    {        int u,v,z;        char ch;        memset(maps,0,sizeof(maps));        for(int i=0; i<m; i++)        {            ch=getchar();            while(ch!='(')                ch=getchar();            scanf("%d,%d)%d",&u,&v,&z);            maps[u+1][v+1]+=z;    //给出点的从0开始，所有的点后移一个，腾出0做超级源点        }        for(int i=0; i<np; i++)        {            ch=getchar();            while(ch!='(')                ch=getchar();            scanf("%d)%d",&u,&z);            maps[0][u+1]+=z;     //所有发电站与超级源点相连        }        for(int i=0; i<nc; i++)        {            ch=getchar();            while(ch!='(')                ch=getchar();            scanf("%d)%d",&u,&z);            maps[u+1][n+1]+=z;    //n+1点作为超级汇点，所有的消耗站与超级汇点相连        }        printf("%d\n",EK_Max_flow(0,n+1));    }    return 0;}