HDU 3157 Crazy Circuits

题目链接

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,每一个线路有一个下限值求一个能够让全部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。然后增加t->s,容量INF的边,在跑一次ss->tt,假设是满流。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增加直接跑一下。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;
}

04-25 16:56