求无向图的点连通度,一般的方法就是转化为网络流来求解
构建网络流模型:
若G为无向图:
(1)原G图中的每个顶点V变成N网中的两个顶点V'和V'',顶点V'至V''有一条弧容量为1;
(2)原图G中的每条边e=(U,V),在N网中有两条弧e'=(U'',V'),e''=(V'',U')与之对应,e'与e''容量均为无穷;
(3)以某点为源点,枚举汇点,求最大流。
其中源点的确立有一个条件,就是这个点不能和其他的所有点都相邻,如果都相邻,显然是无法求出最小割的。 所以在确立源点的时候不能直接随意选点。
转载了一段关于为什么只要枚举汇点:
假设点连通度为k,那么如果你枚举源汇,我们构图的时候是拆点的,所有最小割中最小的一定是k,对应k个点,割边为这k个点拆点后对应的边,那么,这个最小割把所有点点分成了两部分,S和T集合,S和T集合中的点都是不连通的,而S集合中的点都是连通的,T集合中的点也都是连通的,对于任意一个点i属于S,任意一个点j属于T,要使得他们不连通,在图中删除的点都为k,不可能更小了,否则最小割值比k小,而如果他们同时属于S集合或者T集合,使得他们不连通,要删除的点大于k,因为他们现在还连通,于是,如果我们指定一个源点,枚举汇点,如果这个源点与汇点同时在刚才那个最小割的S或T集合中,要使他们不连通,删除的点必然大于k,如果他们一个属于S,一个属于T,那么使他们不连通,要删除的点就是k个,所以,只要枚举汇点就行了
/*
ID: sdj22251
PROG: subset
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define LOCA
#define MAXN 1005
#define INF 100000000
#define eps 1e-7
using namespace std;
const int maxnode = 102;
const int infinity = 100000000;
struct edge
{
int ver; // vertex
int cap; // capacity
int flow; // current flow in this arc
edge *next; // next arc
edge *rev; // reverse arc
edge() {}
edge(int Vertex, int Capacity, edge *Next)
:ver(Vertex), cap(Capacity), flow(0), next(Next), rev((edge*)NULL) {}
void* operator new(size_t, void *p) //返回新的空间
{
return p;
}
}*Net[maxnode];
int dist[maxnode]= {0}, numbs[maxnode] = {0}, src, des, n;
void rev_BFS()
{
int Q[maxnode], head = 0, tail = 0;
for(int i = 1; i <= n; ++i)
{
dist[i] = maxnode;
numbs[i] = 0;
}
Q[tail++] = des;
dist[des] = 0;
numbs[0] = 1;
while(head != tail)
{
int v = Q[head++];
for(edge *e = Net[v]; e; e = e->next)
{
if(e -> rev -> cap == 0 || dist[e -> ver] < maxnode)continue;
dist[e -> ver] = dist[v] + 1;
++numbs[dist[e -> ver]];
Q[tail++] = e -> ver;
}
}
}
int maxflow()
{
int u, totalflow = 0;
edge *CurEdge[maxnode], *revpath[maxnode];
for(int i = 1; i <= n; ++i)
{
CurEdge[i]= Net[i];
}
u = src;
while(dist[src] < n)
{
if(u == des) // find an augmenting path
{
int augflow = infinity;
for(int i = src; i != des; i = CurEdge[i] -> ver)
augflow = min(augflow, CurEdge[i] -> cap);
for(int i = src; i != des; i = CurEdge[i] -> ver)
{
CurEdge[i] -> cap -= augflow;
CurEdge[i] -> rev -> cap += augflow;
CurEdge[i] -> flow += augflow;
CurEdge[i] -> rev -> flow -= augflow;
}
totalflow += augflow;
u = src;
}
edge *e;
for(e = CurEdge[u]; e; e = e -> next)
if(e -> cap > 0 && dist[u] == dist[e -> ver] + 1)break;
if(e) // find an admissible arc, then Advance
{
CurEdge[u] = e;
revpath[e -> ver] = e -> rev;
u = e -> ver;
}
else // no admissible arc, then relabel this vertex
{
if(0 == (--numbs[dist[u]]))break; // GAP cut, Important!
CurEdge[u] = Net[u];
int mindist = n;
for(edge *te = Net[u]; te; te = te -> next)
if(te -> cap > 0)mindist = min(mindist, dist[te -> ver]);
dist[u] = mindist + 1;
++numbs[dist[u]];
if(u != src)
u = revpath[u] -> ver; // Backtrack
}
}
return totalflow;
}
int main()
{
int m, u, v, w;
int x[1550], y[1550];
int mp[55][55];
while(scanf("%d%d", &n, &m) != EOF)
{
memset(mp, 0, sizeof(mp));
src = n + 1;
for(int i = 1; i <= m; i++)
{
scanf(" (%d,%d)", &u, &v);
u++;
v++;
x[i] = u;
y[i] = v;
mp[u][v] = mp[v][u] = 1;
}
for(int i = 1; i <= n; i++)
{
bool flag = false;
for(int j = i + 1; j <= n; j++)
{
if(mp[i][j] == 0)
{
flag = true;
break;
}
}
if(flag)
{
src = n + i;
break;
}
}
int N = n;
n = n * 2;
int ans = INF;
for(int j = 1; j <= N; j++)
{
if(src - N == j) continue;
des = j;
memset(Net, 0, sizeof(Net));
memset(dist, 0, sizeof(dist));
memset(numbs, 0, sizeof(numbs));
edge *buffer = new edge[5 * m + 4 * N];
edge *data = buffer;
for(int i = 1; i <= N; i++)
{
u = i;
v = i + N;
w = 1;
Net[u] = new((void*) data++) edge(v, w, Net[u]);
Net[v] = new((void*) data++) edge(u, 0, Net[v]);
Net[u] -> rev = Net[v];
Net[v] -> rev = Net[u];
}
for(int i = 1; i <= m; i++)
{
u = x[i];
v = y[i];
int uu = u + N;
int vv = v + N;
w = infinity;
Net[uu] = new((void*) data++) edge(v, w, Net[uu]);
Net[v] = new((void*) data++) edge(uu, 0, Net[v]);
Net[uu] -> rev = Net[v];
Net[v] -> rev = Net[uu];
Net[vv] = new((void*) data++) edge(u, w, Net[vv]);
Net[u] = new((void*) data++) edge(vv, 0, Net[u]);
Net[vv] -> rev = Net[u];
Net[u] -> rev = Net[vv];
}
rev_BFS();
ans = min(ans, maxflow());
//printf("dfds %d\n", ans);
delete [] buffer;
}
if(ans >= INF) ans = N; //就本题而言,一个点孤立的情况也算连通,所以是n,而一般情况下,完全图的点连通度是n-1
printf("%d\n", ans);
}
return 0;
}