[hdu Solution Notice Code

给你一张图,问至少还要连几条边才能使得整个图都联通。

Solution

我们只要用并查集统计联通块个数即可,然后答案就是n - 1 - 联通块个数。

Notice

注意多组数据和初始化。

Code

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define Rep(i, a, b) for (int i = (a); i < (b); i++)
#define travel(i, u) for (int i = head[u]; ~i; i = edge[i].next)

const ll INF = 1e9, Mo = 998244353;
const int N = 30000;
const double eps = 1e-6;
namespace slow_IO
{
ll ()
{
ll x = 0; int zf = 1; char ch = getchar();
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x * zf;
}
void write(ll y)
{
if (y < 0) putchar('-'), y = -y;
if (y > 9) write(y / 10);
putchar(y % 10 + '0');
}
}
using namespace slow_IO;

char st[5];
int n = 0, now = 0;
struct UnionFindSet
{
int Fa[N + 5], Rank[N + 5];
void Make_Set(int x)
{
Fa[x] = x, Rank[x] = 0;
}
int Find(int x)
{
return Fa[x] == x ? x : Fa[x] = Find(Fa[x]);
}
void Union(int x, int y)
{
int fx = Find(x), fy = Find(y);
if (fx == fy) return;
now++;
if (Rank[fx] < Rank[fy]) Fa[fx] = fy;
else Fa[fy] = fx;
if (Rank[fx] == Rank[fy]) Rank[fx]++;
}
}UFS;
int sqz()
{
while (~scanf("%d", &n))
{
if (!n) break;
int m = read(); now = 0;
rep(i, 1, n) UFS.Make_Set(i);
while (m--)
{
int x = read(), y = read();
UFS.Union(x, y);
}
printf("%dn", n - 1 - now);
}
return 0;
}