hdu 3157 Crazy Circuits （有源汇的有上下界的最小流）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3157

You’ve just built a circuit board for your new robot, and now you need to power it. Your robot circuit consists of a number of electrical components that each require a certain amount of current to operate. Every component has a + and a - lead, which are connected on the circuit board at junctions. Current flows through the component from + to - (but note that a component does not "use up" the current: everything that comes in through the + end goes out the - end). The junctions on the board are labeled 1, ..., N, except for two special junctions labeled + and - where the power supply terminals are connected. The + terminal only connects + leads, and the - terminal only connects - leads. All current that enters a junction from the - leads of connected components exits through connected + leads, but you are able to control how much current flows to each connected + lead at every junction (though methods for doing so are beyond the scope of this problem1). Moreover, you know you have assembled the circuit in such a way that there are no feedback loops (components chained in a manner that allows current to flow in a loop).   Figure 1: Examples of two valid circuit diagrams. In (a), all components can be powered along directed paths from the positive terminal to the negative terminal. In (b), components 4 and 6 cannot be powered, since there is no directed path from junction 4 to the negative terminal. In the interest of saving power, and also to ensure that your circuit does not overheat, you would like to use as little current as possible to get your robot to work. What is the smallest amount of current that you need to put through the + terminal (which you can imagine all necessarily leaving through the - terminal) so that every component on your robot receives its required supply of current to function? Hint 1 For those who are electronics-inclined, imagine that you have the ability to adjust the potential on any componentwithout altering its current requirement, or equivalently that there is an accurate variable potentiometer connected in series with each component that you can adjust. Your power supply will have ample potential for the circuit.     Input The input file will contain multiple test cases. Each test case begins with a single line containing two integers: N (0 <= N <= 50), the number of junctions not including the positive and negative terminals, and M (1 <= M <= 200), the number of components in the circuit diagram. The next M lines each contain a description of some component in the diagram. The ith component description contains three fields: pi, the positive junction to which the component is connected, ni, the negative junction to which the component is connected, and an integer Ii (1 <= Ii <= 100), the minimum amount of current required for component i to function. The junctions pi and ni are specified as either the character '+' indicating the positive terminal, the character '-' indicating the negative terminal, or an integer (between 1 and N) indicating one of the numbered junctions. No two components have the same positive junction and the same negative junction. The end-of-file is denoted by an invalid test case with N = M = 0 and should not be processed.     Output For each input test case, your program should print out either a single integer indicating the minimum amount of current that must be supplied at the positive terminal in order to ensure that every component is powered, or the message "impossible" if there is no way to direct a sufficient amount of current to each component simultaneously.     Sample Input   6 10 + 1 1 1 2 1 1 3 2 2 4 5 + - 1 4 3 2 3 5 5 4 6 2 5 - 1 6 5 3 4 6 + 1 8 1 2 4 1 3 5 2 4 6 3 - 1 3 4 3 0 0     Sample Output   9 impossible

 

题意：有两个正负极n个节点和m个元件，每个元件告诉端点是接在哪个节点上的，并且每个元件有工作的最小电流限制，问使所有元件工作的满足条件的最小电流是多少。

流量有上下限的有源汇的最小流建图：

1.u到v建容量为C[u,v]-B[u.v]的边 //C是上界，B是下界
2.对于点u,记tmp为in[u]-ou[u],若tmp>0，则S向u建容量为tmp的边，若tmp<0，则u向T建边，容量是-tmp。//in[u]记录u的入边的流量下界和，ou[u]记录u的出边的流量的下界和

先跑一遍最大流，然后源点和汇点连一条容量为inf的边，再跑一遍最大流，就是解

#pragma GCC optimize(2)
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 5005;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int node, src, dest, edge;
int ver[maxn], flow[maxn], _next[maxn];
int head[maxn], work[maxn], dis[maxn], q[maxn], in[maxn];
void init(int _node, int _src, int _dest)
{
	node = _node, src = _src, dest = _dest;
	for (int i = 0; i < node; ++i)
	{
		head[i] = -1, in[i] = 0;
	}
	edge = 0;
}
void addedge(int u, int v, int c)
{
	ver[edge] = v, flow[edge] = c, _next[edge] = head[u], head[u] = edge++;
	ver[edge] = u, flow[edge] = 0, _next[edge] = head[v], head[v] = edge++;
}
bool Dinic_bfs()
{
	int i, u, v, l, r = 0;
	memset(dis, -1, sizeof(dis));
	queuepq;
	dis[src] = 0;
	pq.push(src);
	while (!pq.empty())
	{
		int u = pq.front();
		pq.pop();
		for (i = head[u]; i >= 0; i = _next[i])
		{
			int v = ver[i];
			if (flow[i] && dis[v] < 0)
			{
				dis[v] = dis[u] + 1;
				pq.push(v);
				if (v == dest)
				{
					return 1;
				}
			}
		}
	}
	return 0;
}
int Dinic_dfs(int u, int f)
{
	if (u == dest)
	{
		return f;
	}
	int v, tmp;
	for (int &i = work[u]; i >= 0; i = _next[i])
	{
		v = ver[i];
		if (flow[i] && dis[v] == dis[u] + 1)
		{
			tmp = Dinic_dfs(v, min(f, flow[i]));
			if (tmp > 0)
			{
				flow[i] -= tmp;
				flow[i ^ 1] += tmp;
				return tmp;
			}
		}
	}
	return 0;
}
void Dinic_flow()
{
	while (Dinic_bfs())
	{
		for (int i = 0; i < node; ++i)
		{
			work[i] = head[i];
		}
		while (Dinic_dfs(src, inf));
	}
}
int Limit_flow()
{
	int i, src0, dest0;
	src0 = src, dest0 = dest;
	src = node++, dest = node++;
	head[src] = head[dest] = -1;
	for (i = 0; i < node - 2; ++i)
	{
		if (in[i] > 0)addedge(src, i, in[i]);
		if (in[i] < 0)addedge(i, dest, -in[i]);
	}
	Dinic_flow();
	addedge(dest0, src0, inf);
	Dinic_flow();
	for (i = head[src]; i >= 0; i = _next[i])
		if (flow[i])return -1;
	for (i = head[dest0]; i >= 0; i = _next[i])
		if (ver[i] == src0)return flow[i ^ 1];
	return 0;
}
void get(int &a)
{
	char c;
	while (((c = getchar()) < '0' || c > '9') && c != '-'&&c != '+');
	if (c == '-') { a = dest; return; }
	if (c == '+') { a = src; return; }
	for (a = 0; c >= '0'&&c <= '9'; c = getchar())a = a * 10 + c - '0';
}
int main()
{
	//freopen("C:/input.txt", "r", stdin);
	int u, v, c, n, m, ans;
	while (get(n), get(m), n + m)
	{
		init(n + 2, 0, n + 1);
		while (m--)
		{
			get(u), get(v), get(c);
			in[u] -= c;
			in[v] += c;
			addedge(u, v, inf);
		}
		if ((ans = Limit_flow()) >= 0)
		{
			printf("%d\n", ans);
		}
		else 
			printf("impossible\n");
	}
}