UVA 11248 - Frequency Hopping（网络流量）

UVA 11248 - Frequency Hopping

题目链接

题意：给定一个网络，如今须要从1到N运输流量C，问是否可能，假设可能输出可能，假设不可能，再问能否通过扩大一条边的容量使得可能，假设能够输出这些边（按u先排再按v排），假设不行输出不可能

思路：先做一遍网络流，然后每次在最小割上进行添加容量，须要两个优化，每次找流量找到>= c就能够了，然后每次改动容量，能够直接从之前做过的网络流继续做就可以

代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;

const int MAXNODE = 105 * 2;
const int MAXEDGE = 100005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

int n, m, c;

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;
}
};

bool cmp(Edge a, Edge b) {
if (a.u != b.u) return a.u < b.u;
return a.v < b.v;
}

struct Dinic {
int n, m, s, t;
Edge edges[MAXEDGE], etmp[MAXEDGE];
int first[MAXNODE];
int next[MAXEDGE];
bool vis[MAXNODE];
Type d[MAXNODE];
int cur[MAXNODE];
vector<int> cut;
vector<Edge> ans;
Type flow;

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;
}

void MinCut() {
cut.clear();
for (int i = 0; i < m; i += 2) {
if (vis[edges[i].u] && !vis[edges[i].v])
cut.push_back(i);
}
}

bool 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);
if (flow >= c) return true;
}
return false;
}

void solve() {
if (Maxflow(0, n - 1) || c == 0) printf("possible\n");
else {
MinCut();
ans.clear();
for (int i = 0; i < m; i++)
etmp[i] = edges[i];
int tmpf = flow;
for (int i = 0; i < cut.size(); i++) {
edges[cut[i]].cap = edges[cut[i]].flow + c;
if (Maxflow(0, n - 1)) ans.push_back(edges[cut[i]]);
flow = tmpf;
for (int i = 0; i < m; i++)
edges[i] = etmp[i];
}
if (ans.size() == 0) printf("not possible\n");
else {
sort(ans.begin(), ans.end(), cmp);
printf("possible option:");
for (int i = 0; i < ans.size(); i++)
printf("(%d,%d)%c", ans[i].u + 1, ans[i].v + 1, i == ans.size() - 1 ? '\n' : ',');
}
}
}
} gao;

int main() {
int cas = 0;
while (~scanf("%d%d%d", &n, &m, &c) && n) {
gao.init(n);
int u, v, cap;
while (m--) {
scanf("%d%d%d", &u, &v, &cap);
u--; v--;
gao.add_Edge(u, v, cap);
}
printf("Case %d: ", ++cas);
gao.solve();
}
return 0;
}