poj1459——Power Network（網路流，超級源點、匯點）

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)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

題意比較難懂。而且很多廢話，直接看輸入。先是四個數n,np,nc,m，n表示節點，有發電廠，變壓器，使用者等等；接下來是m對數，括弧裡是路徑，然後就是流量；接下來是np個數，括弧裡是發電廠的位置，外面是它的容量；最後是nc個數，表示使用者的位置和容量。
容易看出發電廠和使用者的個數有很多個，所以要建立超級源點串連所有發電廠，發電廠的容量表示路徑的流量。同理用超級匯點串連所有使用者，使用者的容量表示路徑的流量。接著用最大流解決

#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <vector>#include <iostream>#include <set>#include <cstring>#include <string>#define MAXN 510#define inf 0x3f3f3f3fusing namespace std;int n,np,nc,m;int flow[MAXN][MAXN],cap[MAXN][MAXN],pre[MAXN],a[MAXN]; //流量，容量，父節點，殘量void EdmondsKarp(int s,int t){    int u,v;    queue<int> q;    int f=0;    while(1)    {        memset(a,0,sizeof(a));        q.push(s);        a[s]=inf;        while(!q.empty()) //bfs找增光路徑        {            u=q.front();            q.pop();            for(v=0; v<=n+1; ++v)            {                if(a[v]==0&&cap[u][v]>flow[u][v])                {                    pre[v]=u;                    a[v]=min(a[u],cap[u][v]-flow[u][v]); //找到s-v路徑的最小殘量                    q.push(v);                }            }        }        if(a[t]==0) //所有路徑的殘量都為0，說明已經是最大流            break;        for(v=t; v!=s; v=pre[v]) //更新路徑的流量        {            flow[pre[v]][v]+=a[t];            flow[v][pre[v]]-=a[t];        }        f+=a[t];    }    printf("%d\n",f);}int main(){    int u,v,w;    char ch;    while(~scanf("%d%d%d%d",&n,&np,&nc,&m))    {        memset(cap,0,sizeof(cap));        memset(flow,0,sizeof(flow));        for(int i=0; i<m; ++i) //輸入路徑，要預留超級源點的位置        {            cin>>ch>>u>>ch>>v>>ch>>w;            cap[u+1][v+1]=w;        }        for(int i=0;i<np;++i)        {            cin>>ch>>u>>ch>>w; //發電廠的位置，從超級源點0開始            cap[0][u+1]=w;        }        for(int i=0;i<nc;++i) //使用者的位置，到達超級匯點n+1        {            cin>>ch>>u>>ch>>w;            cap[u+1][n+1]=w;        }        EdmondsKarp(0,n+1);    }    return 0;}