链接：http://poj.org/problem?id=1966
题意：有线电视网络中，中继器的连接是双向的。如果网络中任何两个中继器之间至少有一条路，则中继器网络称为是连通的，否则中继器网络是不连通的。一个空的网络、以及只有一个中继器的网络被认为是连通的。具有n 个中继器的网络的安全系数f 被定义成：
(1) f 为n，如果不管删除多少个中继器，剩下的网络仍然是连通的；
(2) f 为删除最少的顶点数，使得剩下的网络不连通。
现在给定一个有线电视网络，求 f 为多少。
思路：一张连通图，求最少删去多少个点，能够使得该图不再连通，这是无向图的顶点连通度问题。
无向图的点连通度：
1. 将每个点u拆为u‘, u‘‘.顶点u‘到u‘‘连一条弧，容量为1。
2. 将图中的每条边(u, v)拆成<u‘‘, v‘>和<v‘‘, u‘>两条边，每条边的容量为INF。
3. 选一个源点A‘‘, 枚举汇点B‘. 求出最大流的最小值即为点连通度。
4. 所有具有流量1的弧(v‘, v‘‘)对应的顶点v构成了一个割点集。
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define per(i, a, n) for (int i = n - 1; i >= a; i--)
#define eps 1e-6
#define debug puts("===============")
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define POSIN(x,y) (0 <= (x) && (x) < n && 0 <= (y) && (y) < m)
typedef long long ll;
typedef unsigned long long ULL;
const int maxn = 1000;
const int maxm = 100000;
int edge[maxm][2];
struct node {
int v; // vertex
int cap; // capacity
int flow; // current flow in this arc
int nxt;
} e[maxm * 2];
int g[maxn], cnt;
int st, ed, n, m;
void add(int u, int v, int c) {
e[++cnt].v = v;
e[cnt].cap = c;
e[cnt].flow = 0;
e[cnt].nxt = g[u];
g[u] = cnt;
e[++cnt].v = u;
e[cnt].cap = 0;
e[cnt].flow = 0;
e[cnt].nxt = g[v];
g[v] = cnt;
}
void init(int N) {
cnt = 1;
st = N;
for (int i = 0; i < n; i++) g[i] = 0;
int u, v;
for (int i = 0; i < N; i++) add(i, i + N, 1);
for (int i = 0; i < m; i++) {
int u = edge[i][0], v = edge[i][1];
add(v + N, u, INF);
add(u + N, v, INF);
}
}
int dist[maxn], numbs[maxn], q[maxn];
void rev_bfs() {
int font = 0, rear = 1;
for (int i = 0; i <= n; i++) { //n为总点数
dist[i] = maxn;
numbs[i] = 0;
}
q[font] = ed;
dist[ed] = 0;
numbs[0] = 1;
while(font != rear) {
int u = q[font++];
for (int i = g[u]; i; i = e[i].nxt) {
if (e[i ^ 1].cap == 0 || dist[e[i].v] < maxn) continue;
dist[e[i].v] = dist[u] + 1;
++numbs[dist[e[i].v]];
q[rear++] = e[i].v;
}
}
}
int maxflow() {
rev_bfs();
int u, totalflow = 0;
int curg[maxn], revpath[maxn];
for(int i = 0; i <= n; ++i) curg[i] = g[i];
u = st;
while(dist[st] < n) {
if(u == ed) { // find an augmenting path
int augflow = INF;
for(int i = st; i != ed; i = e[curg[i]].v)
augflow = min(augflow, e[curg[i]].cap);
for(int i = st; i != ed; i = e[curg[i]].v) {
e[curg[i]].cap -= augflow;
e[curg[i] ^ 1].cap += augflow;
e[curg[i]].flow += augflow;
e[curg[i] ^ 1].flow -= augflow;
}
totalflow += augflow;
u = st;
}
int i;
for(i = curg[u]; i; i = e[i].nxt)
if(e[i].cap > 0 && dist[u] == dist[e[i].v] + 1) break;
if(i) { // find an admissible arc, then Advance
curg[u] = i;
revpath[e[i].v] = i ^ 1;
u = e[i].v;
} else { // no admissible arc, then relabel this vertex
if(0 == (--numbs[dist[u]])) break; // GAP cut, Important!
curg[u] = g[u];
int mindist = n;
for(int j = g[u]; j; j = e[j].nxt)
if(e[j].cap > 0) mindist = min(mindist, dist[e[j].v]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != st)
u = e[revpath[u]].v; // Backtrack
}
}
return totalflow;
}
int main () {
int N;
while(~scanf("%d%d", &N, &m)) {
n = 2 * N + 4;
for (int i = 0; i < m; i++) scanf(" (%d,%d)", edge[i], edge[i] + 1);
int ans = INF;
for (int i = 1; i < N; i++) {
ed = i;
init(N);
ans = min(ans, maxflow());
}
if (ans == INF) ans = N;
printf("%d\n", ans);
}
return 0;
}POJ 1966 Cable TV Network（无向图的顶点连通度）
原文地址：http://blog.csdn.net/sio__five/article/details/38961679
