您的位置:首页 > Web前端

POJ 3155 Hard Life(最大密度子图)

2017-04-26 09:26 411 查看

POJ 3155 Hard Life

题目链接

最大密度子图模板题

代码:

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

const int MAXNODE = 1105;
const int MAXEDGE = 100005;

typedef double Type;
const Type INF = 1e15;

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];
int cur[MAXNODE];
vector<int> cut;

void init(int n) {
this->n = n;
memset(first, -1, sizeof(first));
m = 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;
Type flow = 0;
while (bfs()) {
for (int i = 0; i < n; i++)
cur[i] = first[i];
flow += dfs(s, INF);
}
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);
}
}
} gao;

const int N = 1005;
const double eps = 1e-8;

int n, m, u
, v
, s, t, du
, sum;

double build(double g) {
gao.init(n + 2);
for (int i = 1; i <= n; i++) {
gao.add_Edge(s, i, 1.0 * m);
gao.add_Edge(i, t, m + 2 * g - du[i]);
}
for (int i = 0; i < m; i++) {
gao.add_Edge(u[i], v[i], 1.0);
gao.add_Edge(v[i], u[i], 1.0);
}
return n * m - gao.Maxflow(s, t);
}

void dfs(int u) {
gao.vis[u] = true;
if (u >= 1 && u <= n) sum++;
for (int i = gao.first[u]; i + 1; i = gao.next[i]) {
int v = gao.edges[i].v;
if (gao.edges[i].flow < gao.edges[i].cap && !gao.vis[v]) dfs(v);
}
}

int main() {
while (~scanf("%d%d", &n, &m)) {
if (m == 0) {
printf("1\n1\n");
continue;
}
s = 0, t = n + 1;
memset(du, 0, sizeof(du));
for (int i = 0; i < m; i++) {
scanf("%d%d", &u[i], &v[i]);
du[u[i]]++; du[v[i]]++;
}
double l = 0, r = m;
while (r - l >= 1.0 / n / n) {
double mid = (l + r) / 2;
if (build(mid) < eps) r = mid;
else l = mid;
}
build(l);
gao.Maxflow(s, t);
memset(gao.vis, false, sizeof(gao.vis));
sum = 0;
dfs(s);
printf("%d\n", sum);
for (int i = 1; i <= n; i++)
if (gao.vis[i]) printf("%d\n", i);
}
return 0;
}


内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: