1821: [JSOI2010]Group 部落划分 Group
分析：
二分，然后把小于这条边的全连上，然后判断联通块的个数，如果<k，那么说明mid太大，否则说明mid太小。
其实有更奇妙的思路，从小的往大的加边，一旦加到使联通块的个数==k-1了，说明k个联通块已经构造出来了，再加入这一条就不满足了，这一条边就是答案。
代码：
二分
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 inline int read() { 6 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 7 for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 8 } 9 10 const double eps = 1e-4; 11 const int N = 1010; 12 struct Edge{ 13 int u,v; 14 double w; 15 Edge() {} 16 Edge(int a,int b,double c) {u = a, v = b, w = c;} // -- double c 17 bool operator < (const Edge &A) const { 18 return w < A.w; 19 } 20 }e[N*N]; 21 int x[N],y[N],fa[N],Enum,n,k; 22 23 int find(int x) { 24 if (x == fa[x]) return x; 25 return fa[x] = find(fa[x]); 26 } 27 bool check(double x) { 28 for (int i=1; i<=n; ++i) fa[i] = i; 29 int cnt = 0; 30 for (int i=1; i<=Enum; ++i) { // -- i<=n 31 if (e[i].w > x) break; 32 int u = find(e[i].u),v = find(e[i].v); 33 if (u != v) { 34 fa[u] = v; 35 cnt ++; 36 if (cnt == n - 1) break; 37 } 38 } 39 if (n - cnt < k) return false; // 联通块的个数 40 return true; 41 } 42 int main() { 43 n = read(),k = read(); 44 double L = 0,R = 0,mid,ans; 45 for (int i=1; i<=n; ++i) 46 x[i] = read(),y[i] = read(); 47 for (int i=1; i<=n; ++i) 48 for (int j=i+1; j<=n; ++j) { 49 double w = sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j])); 50 e[++Enum] = Edge(i,j,w); 51 R = max(R,w); 52 } 53 sort(e+1,e+Enum+1); 54 while (R - L >= eps) { 55 mid = (L + R) / 2.0; 56 if (check(mid)) ans = mid,L = mid; // -- L = mid + 1 57 else R = mid; // -- R = mid - 1 58 } 59 printf("%.2lf",ans); 60 return 0; 61 }
另一种奇妙的写法。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 5 inline int read() { 6 int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; 7 for (;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; 8 } 9 10 const int N = 1010; 11 struct Edge{ 12 int u,v;double w; 13 Edge() {} 14 Edge(int a,int b,double c) {u = a, v = b, w = c;} // double c 15 bool operator < (const Edge &A) const { 16 return w < A.w; 17 } 18 }e[N*N]; 19 int x[N],y[N],fa[N],Enum,n,k; 20 21 int find(int x) { 22 if (x == fa[x]) return x; 23 return fa[x] = find(fa[x]); 24 } 25 int main() { 26 n = read(),k = read(); 27 for (int i=1; i<=n; ++i) 28 x[i] = read(),y[i] = read(); 29 for (int i=1; i<=n; ++i) 30 for (int j=i+1; j<=n; ++j) { 31 double w = sqrt(1.0*(x[i]-x[j])*(x[i]-x[j])+1.0*(y[i]-y[j])*(y[i]-y[j])); 32 e[++Enum] = Edge(i,j,w); 33 } 34 sort(e+1,e+Enum+1); 35 for (int i=1; i<=n; ++i) fa[i] = i; 36 int cnt = n; 37 for (int i=1; i<=Enum; ++i) { // -- i<=n 38 int u = find(e[i].u), v = find(e[i].v); 39 if (u != v) { 40 fa[u] = v; 41 cnt --; 42 } 43 if (cnt == k - 1) {printf("%.2lf",e[i].w);return 0;} 44 } 45 return 0; 46 }