对两个连通块做最小生成树,然后中间的边选一条代价最小的就行了。。。。
#include <iostream> #include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits> #include <cstdlib> #include <cmath> #include <time.h> #define maxn 1005 #define maxm 1000005 #define eps 1e-7 #define mod 1000000007 #define INF 0x3f3f3f3f #define PI (acos(-1.0)) #define lowbit(x) (x&(-x)) #define mp make_pair #define ls o<<1 #define rs o<<1 | 1 #define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R #define pii pair<int, int> #pragma comment(linker, "/STACK:16777216") typedef long long LL; typedef unsigned long long ULL; //typedef int LL; using namespace std; LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;} LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;} // head struct node { int u, v; double c; node(int u = 0, int v = 0, double c = 0) : u(u), v(v), c(c) {} }e[maxm]; int flag[maxn]; int f[maxn]; pair<double, double> p[maxn]; double tt[maxm]; int cnt, n; int find(int u) { return f[u] = f[u] == u ? f[u] : find(f[u]); } int cmp(node aa, node bb) { return aa.c < bb.c; } bool merge(int u, int v) { int uu = find(u), vv = find(v); if(uu == vv) return false; f[uu] = vv; return true; } void work() { int kk, d; double u, v; scanf("%d%d", &d, &n); cnt = 0; for(int i = 0; i < n; i++) { scanf("%lf%lf%d", &u, &v, &kk); p[i] = mp(u, v); flag[i] = kk; } int tcnt = 0; double res = 1e15; for(int i = 0; i < n; i++) for(int j = i+1; j < n; j++) { double t1 = p[i].first - p[j].first; double t2 = p[i].second - p[j].second; double t = t1 * t1 + t2 * t2; t = sqrt(t); if(flag[i] ^ flag[j]) res = min(res, t); else e[cnt++] = node(i, j, t); } if(fabs(res - 1e15) > eps) tt[tcnt++] = res; sort(e, e+cnt, cmp); for(int i = 0; i < n; i++) f[i] = i; for(int i = 0; i < cnt; i++) { if(merge(e[i].u, e[i].v)) tt[tcnt++] = e[i].c; } double ans = 0; sort(tt, tt+tcnt); for(int i = 0; i < tcnt - d; i++) ans += tt[i]; printf("%.2f\n", ans); } int main() { int _; scanf("%d", &_); while(_--) work(); return 0; }