1821: [JSOI2010]Group 部落划分 Group

1821: [JSOI2010]Group 部落划分 Group

链接

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 }

View Code

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 }

View Code