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题意:一个电路板,上面有N个接线柱(标号1~N),还有两个电源接线柱 + -,给出一些线路,每个线路有一个下限值求一个可以让所有部件正常工作的总电流 没有则输出impossible
思路:
有源汇有上下界求最小流,建模方法为:
按无源汇先建图,跑超级源汇ss->tt一次,然后加入t->s,容量INF的边,在跑一次ss->tt,如果是满流,就有解,解为t->s边的当前流量
顺带写个最大流的,最大流就先把t->s加入直接跑一下,t->s的流量就是了
代码:
#include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int MAXNODE = 65; const int MAXEDGE = 10005; typedef int Type; const Type INF = 0x3f3f3f3f; struct Edge { int u, v; Type cap, flow; Edge() {} Edge(int u, int v, Type cap, Type flow) { this->u = u; this->v = v; this->cap = cap; this->flow = flow; } }; struct Dinic { int n, m, s, t; Edge edges[MAXEDGE]; int first[MAXNODE]; int next[MAXEDGE]; bool vis[MAXNODE]; Type d[MAXNODE]; Type flow; int cur[MAXNODE]; void init(int n) { this->n = n; memset(first, -1, sizeof(first)); m = 0; flow = 0; } void add_Edge(int u, int v, Type cap) { edges[m] = Edge(u, v, cap, 0); next[m] = first[u]; first[u] = m++; edges[m] = Edge(v, u, 0, 0); next[m] = first[v]; first[v] = m++; } bool bfs() { memset(vis, false, sizeof(vis)); queue<int> Q; Q.push(s); d[s] = 0; vis[s] = true; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = first[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (!vis[e.v] && e.cap > e.flow) { vis[e.v] = true; d[e.v] = d[u] + 1; Q.push(e.v); } } } return vis[t]; } Type dfs(int u, Type a) { if (u == t || a == 0) return a; Type flow = 0, f; for (int &i = cur[u]; i != -1; i = next[i]) { Edge& e = edges[i]; if (d[u] + 1 == d[e.v] && (f = dfs(e.v, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[i^1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } Type Maxflow(int s, int t) { this->s = s; this->t = t; while (bfs()) { for (int i = 0; i < n; i++) cur[i] = first[i]; flow += dfs(s, INF); } return flow; } bool judge(int s) { for (int i = first[s]; i + 1; i = next[i]) if (edges[i].flow != edges[i].cap) return false; return true; } } gao; const int N = 65; int n, m, s, t, ss, tt, du[N]; char c[15]; int main() { while (~scanf("%d%d", &n, &m) && n || m) { gao.init(n + 4); s = 0; t = n + 1; ss = n + 2; tt = n + 3; int u, v, w; memset(du, 0, sizeof(du)); while (m--) { scanf("%s", c); if (c[0] == '+') u = s; else sscanf(c, "%d", &u); scanf("%s", c); if (c[0] == '-') v = t; else sscanf(c, "%d", &v); scanf("%d", &w); gao.add_Edge(u, v, INF); du[u] -= w; du[v] += w; } for (int i = s; i <= t; i++) { if (du[i] > 0) gao.add_Edge(ss, i, du[i]); if (du[i] < 0) gao.add_Edge(i, tt, -du[i]); } gao.Maxflow(ss, tt); gao.add_Edge(t, s, INF); gao.Maxflow(ss, tt); if (!gao.judge(ss)) printf("impossible\n"); else printf("%d\n", gao.edges[gao.m - 2].flow); } return 0; }
HDU 3157 Crazy Circuits(有源汇上下界最小流)
标签:style blog http io color ar os sp for
原文地址:http://blog.csdn.net/accelerator_/article/details/41051771