UVALive 7456 Least Crucial Node

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5478



题意：有n个点，m条边，给出点S（1~n中的某个点）。除了S点之外的n-1个点中的一个如果失效，那么可能会影响到若干个点和S的连通性。现在要找一个点p，使得p失效后使得点和S不连通的数量最大，在此基础上p的标号越小。



思路：枚举失效n-1个点，每次做一遍并查集，检查有多少点会失效。然后取最大值即可。



#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <utility>
using namespace std;

#define rep(i,j,k) for (int i=j;i<=k;i++)
#define Rrep(i,j,k) for (int i=j;i>=k;i--)

#define Clean(x,y) memset(x,y,sizeof(x))
#define LL long long
#define ULL unsigned long long
#define inf 0x7fffffff
#define mod 100000007

#define mp make_pair
#define fi first
#define se second

#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m=(l+r)>>1
const int maxn = 110;

int n,m,st;

int x[maxn*maxn];
int y[maxn*maxn];
int k[maxn];

int father[maxn];

int getfather(int x)
{
return x==father[x]?x:father[x] = getfather(father[x]);
}

void init()
{
scanf("%d",&st);
scanf("%d",&m);
rep(i,1,m) scanf("%d%d",&x[i],&y[i]);
Clean(k,0);
}

void unio( int a , int b )
{
int fa = getfather(a);
int fb = getfather(b);
if ( fa != fb )
{
father[fa] = fb;
}
}
void solve()
{
rep(i,1,n)
{
if ( i == st ) continue;
rep(j,1,n) father[j] = j;
rep(j,1,m)
if ( x[j] != i && y[j] != i )
unio( x[j] , y[j] );
rep(j,1,n)
if ( j != i && getfather(j) != getfather(st) ) k[i]++;
}
int ans,M = 0;
rep(i,1,n)
if ( k[i] > M )
{
M = k[i];
ans = i;
}
printf("%d\n",ans);
}

int main()
{
while( scanf("%d",&n) == 1 )
{
if ( !n ) break;
init();
solve();
}
return 0;
}