Power Network(最大流基礎_增廣路演算法：多源多匯，自建總源點和總匯點)

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Power Network Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64uSubmitStatus

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

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.
ps：坑我的一道題啊，題目看了好長時間才明白，那麼多資料直接嚇哭我了，輸入的時候也調試了好久，以後再也不在輸入的時候隨便打空格了。怎麼會有這樣的呢！輸出按ctrl+z
題意：有n個結點，np個發電站，nc個消費者，m個電力運輸線。接下去是n條邊的資訊(u,v)cost，cost表示邊(u,v)的最大流量；a個發電站的資訊(u)cost，cost表示發電站u能提供的最大流量；b個使用者的資訊(v)cost，cost表示每個使用者v能接受的最大流量。思路：在圖中添加1個源點S和匯點T，將S和每個發電站相連，邊的權值是發電站能提供的最大流量；將每個使用者和T相連，邊的權值是每個使用者能接受的最大流量。從而轉化成了一般的最大網路流問題，然後求解。（自己語言總結不好，覺得這個能看懂，就貼一下）

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <algorithm>#include <iostream>#include <queue>using namespace std;#define inf  9999999999int flow[210][210];int maxflow[210],father[210],vis[210];int max_flow;int m,i;void EK(int s,int e){    queue<int >q;    int u,v;    max_flow=0;    while(1)    {        memset(maxflow,0,sizeof(maxflow));        memset(vis,0,sizeof(vis));        maxflow[s]=inf;        q.push(s);        while(!q.empty())        {            u=q.front();            q.pop();            for(v=s;v<=e;v++)            {                if(!vis[v]&&flow[u][v]>0)                {                    vis[v]=1;                    father[v]=u;                    q.push(v);                    maxflow[v]=min(maxflow[u],flow[u][v]);                }            }            if(maxflow[e]>0)            {                while(!q.empty())                    q.pop();                break;            }        }        if(maxflow[e]==0)            break;        for(i=e;i!=s;i=father[i])        {            flow[father[i]][i]-=maxflow[e];            flow[i][father[i]]+=maxflow[e];        }        max_flow+=maxflow[e];    }}int main(){    int n,np,nc,m;int i,a,b,c;char ch;while(~scanf("%d%d%d%d ",&n,&np,&nc,&m))//輸入注意後面的空格{memset(flow,0,sizeof(flow));for(i=1;i<=m;i++){scanf("(%d,%d)",&a,&b);scanf("%d ",&c);//注意後面的空格flow[a+1][b+1]=c;}for(i=1;i<=np;i++)//建立總源點{scanf("(%d)%d ",&a,&b);//注意後面空格flow[0][a+1]=b;}for(i=1;i<=nc;i++)//建立總匯點{            scanf("(%d)%d ",&a,&b);//注意後面的空格flow[a+1][n+1]=b;}EK(0,n+1);printf("%d\n",max_flow);}return 0;}