POJ 1459 Power Network

链接：http://poj.org/problem?id=1459

题目：

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 21928		Accepted: 11453

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)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

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.

解题思路：

这是一道网络流中最大流类型的题目，比较简单，属于入门级，基本上敲一遍模板就可以过了。要注意的一点是:这道题目没有给你源点和汇点，需要自己在构图时添加源点和汇点。由于题目中给的点是从0~n，我的做法是将题中所给的这些点进行偏移，依次加1，这样的话，点的范围就变成了1~n+1，然后，我把0作为源点，n+2作为汇点，用ek算法，求出最大流即可。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;

#define min(x,y) x < y ? x : y
const int MAXN = 105;
int a[MAXN], p[MAXN], cap[MAXN][MAXN], flow[MAXN][MAXN];
int n, np, nc, m;

int ek(int s, int t)
{
	int f = 0;
	memset(p, 0, sizeof(p));
	memset(flow, 0, sizeof(flow));
	queue q;
	
	while(true)
	{
		memset(a, 0, sizeof(a));
		q.push(s);
		a[s] = INT_MAX;
		while(!q.empty())
		{
			int u = q.front(); q.pop();
			for(int v = 0; v <= n+2; v++)
			{
				if(!a[v] && cap[u][v] > flow[u][v])
				{
					p[v] = u; q.push(v);
					a[v] = min(a[u], cap[u][v] - flow[u][v]);
				}
			}
		}
		if(0 == a[t]) break;
		for(int u = t; u != s; u = p[u])
		{
			flow[p[u]][u] += a[t];
			flow[u][p[u]] -= a[t];
		}
		f += a[t];
	}
	
	return f;
}

int main()
{
	while(~scanf("%d%d%d%d", &n, &np, &nc, &m))
	{
		memset(cap, 0, sizeof(cap));
		for(int i = 0; i < m; i++)
		{
			int b, c, d;
			scanf(" (%d,%d)%d", &b, &c, &d);
			cap[b+1][c+1] = d;
		}
		for(int i = 0; i < np; i++)
		{
			int b, c;
			scanf(" (%d)%d", &b, &c);
			cap[0][b+1] = c;
		}
		for(int i = 0; i < nc; i++)
		{
			int b, c;
			scanf(" (%d)%d", &b, &c);
			cap[b+1][n+2] = c;
		}
		printf("%d\n", ek(0, n+2));
	}
	
	return 0;
}