给一个图， n个点m条边， 求至少去掉多少个点可以使得图不再联通。
随便指定一个点为源点， 枚举其他点为汇点的情况， 跑网络流， 求其中最小的情况。 如果最后ans为inf， 说明是一个完全图， 那么结果就为n。
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, a, n) for(int i = a; i<n; i++)
#define ull unsigned long long
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-;
const int mod = 1e9+;
const int inf = ;
const int dir[][] = { {-, }, {, }, {, -}, {, } };
const int maxn = 4e4+;
int q[maxn*], head[maxn*], dis[maxn/], s, t, num, edge[][];
struct node
{
int to, nextt, c;
node(){}
node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
}e[maxn*];
void init() {
num = ;
mem1(head);
}
void add(int u, int v, int c) {
e[num] = node(v, head[u], c); head[u] = num++;
e[num] = node(u, head[v], ); head[v] = num++;
}
int bfs() {
mem(dis);
dis[s] = ;
int st = , ed = ;
q[ed++] = s;
while(st<ed) {
int u = q[st++];
for(int i = head[u]; ~i; i = e[i].nextt) {
int v = e[i].to;
if(!dis[v]&&e[i].c) {
dis[v] = dis[u]+;
if(v == t)
return ;
q[ed++] = v;
}
}
}
return ;
}
int dfs(int u, int limit) {
if(u == t) {
return limit;
}
int cost = ;
for(int i = head[u]; ~i; i = e[i].nextt) {
int v = e[i].to;
if(e[i].c&&dis[v] == dis[u]+) {
int tmp = dfs(v, min(limit-cost, e[i].c));
if(tmp>) {
e[i].c -= tmp;
e[i^].c += tmp;
cost += tmp;
if(cost == limit)
break;
} else {
dis[v] = -;
}
}
}
return cost;
}
int dinic() {
int ans = ;
while(bfs()) {
ans += dfs(s, inf);
}
return ans;
}
int main()
{
int n, m, x, y;
while(~scanf("%d%d", &n, &m)) {
for(int i = ; i<m; i++) {
scanf(" (%d,%d)", &edge[i][], &edge[i][]);
}
s = n;
int ans = inf;
for(int i = ; i<n; i++) {
t = i;
init();
for(int j = ; j<m; j++) {
int x = edge[j][];
int y = edge[j][];
add(x+n, y, inf);
add(y+n, x, inf);
}
for(int j = ; j<n; j++)
add(j, j+n, );
ans = min(ans, dinic());
}
if(ans == inf)
ans = n;
cout<<ans<<endl;
}
return ;
}