zoj 1914 || poj 2349 Arctic Network【最小生成树 kruskal && prim】
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.
Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.
Input
The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).
Output
For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.
Sample Input
1
2 4
0 100
0 300
0 600
150 750
Sample Output
212.13
题意：有两种不同的通信技术，用卫星通信的两个城市之间可以任意联络，但用无线电通信的城市只能和距离不超过D的城市联系。无线电的传输距离D越大，花费就越多。已知无线电的数目，让求最小的D，使得没有卫星的城市都可以使用无线电。
kruskal:
#include<cstdio>
#include<cstring>
#include<cmath>
#define mem(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a) scanf("%d", &a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define INF 0x3f3f3f
#include<algorithm>
using namespace std;
const int mx = 1010;
int per[mx];
void init()
{
for(int i = 0; i < mx; i++)
per[i] = i;
}
struct node
{
int start, end;
double val;
};
node q[1000100];//之前开的不够大 提交后 Segmentation Fault
bool operator < (node a,node b)
{
return a.val < b.val;
}
int find(int x)
{
return x == per[x] ? x : (per[x] = find(per[x]));
}
bool join(int x,int y)
{
int fx = find(x);
int fy = find(y);
if(fx != fy)
{
per[fx] = fy;
return true;
}
return false;
}
double x[mx], y[mx];
int main(){
int t; Si(t);
Wi(t){
init();
int s, p, i, j, k;
scanf("%d%d", &s, &p);
for(i = 0; i < p; i++)
{
scanf("%lf%lf", &x[i], &y[i]);
}
for(i = 0, k = 0; i < p; i++)
{
for(j = i+1; j < p; j++)
{
q[k].start = i;
q[k].end = j;
q[k].val = sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
k++;
}
}
sort(q, q+k);
int num = 0;
double cost;
for(i = 0; i < k; i++)
{
if(join(q[i].start, q[i].end)) num++;
if(num == p-s)
{
cost = q[i].val; break;
}
}
Pf(cost);
}
return 0;
}
prim：
#include<cstdio>
#include<cstring>
#include<cmath>
#define mem(a, b) memset(a, (b), sizeof(a))
#define Wi(a) while(a--)
#define Si(a) scanf("%d", &a)
#define Pf(a) printf("%.2lf\n", (a))
#define INF 0x3f3f3f
#include<algorithm>
using namespace std;
const int mx = 1010;
double map[mx][mx];
double val[mx], dis[mx];//两个数组都是记录权值的
int vis[mx];
int m, n;
int cmp(const double &a, const double &b)
{
return a > b;
}
void prim()
{
mem(vis, 0);
int i, j, k;
double minn;
int t = 0;
for(i = 1; i <= n; i++)
dis[i] = map[1][i];
vis[1] = 1;
for(i = 1; i < n; i++)
{
k = 1; minn = INF;
for(j = 1; j <= n; j++)
{
if(!vis[j] && minn > dis[j])
{
minn = dis[j];
k = j;
}
}
vis[k] = 1;
val[t++] = dis[k];
for(j = 1; j <= n; j++)
{
if(!vis[j] && dis[j] > map[j][k])
dis[j] = map[j][k];
}
}
sort(val, val+t, cmp);//对val从大到小排序
Pf(val[m-1]); //输出除去卫星连接的边的最大边权值
}
double x[mx],y[mx];
int main()
{
int t; Si(t);
Wi(t){
scanf("%d%d", &m, &n);
int i, j, k;
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
map[i][j] = (i==j?0:INF);
}
for(i = 1; i <= n; i++)
{
scanf("%lf%lf", &x[i], &y[i]);
}
for(i = 1; i <= n; i++)
{
for(j = i+1; j <= n; j++)
{
map[i][j] = map[j][i] = sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
}
}
prim();
}
return 0;
}