【并查集 最小生成树两种模板(MST)】

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define MAXN 105
#define dis(a,b) (sqrt((double)(x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b])))
struct NODE{
int a, b;
double v;
bool operator <(const NODE& t)const {return v < t.v;}
}p[MAXN*MAXN];
int x[MAXN], y[MAXN], f[MAXN], n,m;

int find(int x){return f[x] == x ? x : f[x] = find(f[x]);}
double kruskal()
{
for(int i = 0; i <= n; i++)f[i] = i;
double ans = 0;
int cnt = 1;
sort(p,p+m);
for(int i = 0; i < m; i++)
{
int a = find(p[i].a), b = find(p[i].b);
if(a != b) ans += p[i].v, f[a] = b, cnt++;
if(cnt == n) break;
}
if(cnt < n) return -1;
return ans;
}
int main()
{
#ifdef SHY
freopen("e:\\1.txt","r",stdin);
#endif
int t;
scanf("%d%*c", &t);
while(t--)
{
m = 0;
scanf("%d%*c", &n);
for(int i = 0; i < n; i++)
scanf("%d %d%*c", &x[i], &y[i]);
for(int i = 0; i < n; i++)
{
for(int j = i+1; j < n; j++)
{
p[m].v= dis(i,j);
if(10.0 <= p[m].v && 1000.0 >= p[m].v)
p[m].a = i, p[m++].b = j;
}
}
double ans = kruskal();
if(ans > 0.0)
printf("%.1lf\n", ans*100.0);
else puts("oh!");
}
return 0;
}

#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
typedef long long LL;
#define MAXN 110
int cost[MAXN][MAXN], n, low[MAXN], u[MAXN], v[MAXN], f[MAXN],len;
bool vis[MAXN];
/*f[]记录路径(父亲节点)，*/

int prim()
{
int ans = 0;
len = 0;
clc(vis,0);
vis[1] = true;//从1开始查找
repe(i,2,n)  low[i] = cost[1][i], f[i] = 1;
repe(i,2,n)
{
int mi = INF, p = -1;
repe(j,1,n)//找到没有选择点的最小的距离
{
if(!vis[j] && mi > low[j])
{
mi = low[j];
p = j;
}
}
if(INF == mi) return -1;//原图不连通
ans += mi;
if(mi)//本题要求路径u<v,且0花费路径不需要记录（因为本题0是已经存在的路）
u[len] = min(p,f[p]), v[len++] = max(p,f[p]);
vis[p] = true;
repe(j,1,n)
{
if(!vis[j] && low[j] > cost[p][j])//更新距离和f[]
low[j] = cost[p][j], f[j] = p;
}
}
return ans;
}