POJ

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)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

HintThe 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.

题意：这题不难，但这个破题死活读不懂（建议出成6级阅读），后来在网上找了找发现有一个人给的解释还比较清晰这里借用一下。
简单的说下题意（按输入输出来讲，前面的描述一堆的rubbish，还用来误导人），给你n个点，其中有np个是能提供电力的点，nc个是能消费电力的点，剩下的点(n-np-nc)是中转战即不提供电力也不消费电力，点与点之间是有线路存在的，有m条线路，每条线路有最多运载限定。
前4个数据就是有n个点，np个供电点,nc个消费点,m条线路，接来题目先给出的是m条线路的数据，(起点,终点)最多运载量,然后是np个供电点的数据(供电点)最多供电量，接着就是nc个消费点的数据(消费点)最多消费电量。
题目要我们求出给定的图最大能消费的总电量（就是求最大流）

题解：
最大流基础题，就不多说了。注意输入就行，因为输入前面空格数量随机所以建议用cin，嫌慢的可以加上
std::ios::sync_with_stdio(false);
不会用的看这点击打开链接

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 310;

struct Edge{
	int to,flow,rev;
	Edge(){}
	Edge(int a,int b,int c):to(a),flow(b),rev(c){}
};

vector<Edge> E[MAXN];

inline void Add(int from,int to,int flow){
	E[from].push_back(Edge(to,flow,E[to].size()));
	E[to].push_back(Edge(from,0,E[from].size()-1));
}

int deep[MAXN];
int iter[MAXN];
 
bool BFS(int from,int to){
	memset(deep,-1,sizeof deep);
	deep[from] = 0;
	queue<int> Q;
	Q.push(from);
	while(!Q.empty()){
		int t = Q.front();
		Q.pop();
		for(int i=0 ; i<E[t].size() ; ++i){
			Edge& e = E[t][i];
			if(e.flow > 0 && deep[e.to] == -1){
				deep[e.to] = deep[t] + 1;
				Q.push(e.to);
			}
		}
	}
	return deep[to] != -1;
}
 
int DFS(int from,int to,int flow){
	if(from == to || flow == 0)return flow;
	
	for(int& i=iter[from] ; i<E[from].size() ; ++i){
		Edge& e = E[from][i];
		if(e.flow > 0 && deep[e.to] == deep[from]+1){
			int nowflow = DFS(e.to,to,min(flow,e.flow));
			if(nowflow > 0){
				e.flow -= nowflow;
				E[e.to][e.rev].flow += nowflow;
				return nowflow;
			}
		}
	}
	return 0;
}
 
int Dinic(int from,int to){
	int sumflow = 0;
	while(BFS(from,to)){
		memset(iter,0,sizeof iter);
		int mid;
		while((mid=DFS(from,to,INF)) > 0)sumflow += mid;
	}
	return sumflow;
}

int main(){
        int N,P,C,M;
	char rubbish;
        int first, next;
        int flow;
	while(cin>>N>>P>>C>>M){
            for(int i=0 ; i<M ; ++i){
	        cin >> rubbish;
                cin >> first;
                cin >> rubbish;
                cin >> next;
                cin >> rubbish;
                cin >> flow;
                if (first == next)continue;
                Add(first,next,flow);
	    }
	    for(int i=0 ; i<P ; ++i){
	        cin >> rubbish;
                cin >> first;
                cin >> rubbish;
                cin >> flow;
                Add(MAXN-2,first,flow);
	    }
	    for(int i=0 ; i<C ; ++i){
	        cin >> rubbish;
                cin >> first;
                cin >> rubbish;
                cin >> flow;
                Add(first,MAXN-1,flow);
	    }
	    printf("%d
",Dinic(MAXN-2,MAXN-1));
	    for(int i=0 ; i<MAXN ; ++i)E[i].clear();
	}	
	
	return 0;
}