POJ 2349 Arctic Network(Prim算法)

AI音视频处理：视频分割与语义识别技术解析

最新推荐文章于 2020-09-26 09:22:46 发布

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#network #算法 #output #distance #each #pair

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本文深入探讨了AI音视频处理领域中的关键技术，特别是视频分割与语义识别。通过详细解释这些技术的工作原理、应用案例及实际效果，旨在为读者提供全面的理解和洞察。

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4526		Accepted: 1610

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.
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

Sample Output

212.13

Source

Waterloo local 2002.09.28

//Prim算法

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

const double INF = 1<<30;
const int MAX = 505;

struct point {
   double x, y;
}pnt[MAX];

double dist[MAX][MAX], lowcost[MAX], res[MAX];
bool visited[MAX];
int P, S, cnt;
bool cmp(const double a, const double b)
{
   return a > b;
}

void Prim()
{
   int i, j, k;
   cnt = 0;
   for(i = 1; i <= P; i++)
       lowcost[i] = dist[1][i];
   visited[1] = true;
   for(i = 1; i < P; i++)
   {
       double min = INF;
       k = 1;
       for(j = 2; j <= P; j++)
       {
           if((!visited[j]) && (lowcost[j] < min))
           {
               min = lowcost[j];
               k = j;
           }
       }
       res[++cnt] = min;
       visited[k] = true;
       for(j = 2; j <= P; j++)
       {
           if((!visited[j]) && (lowcost[j] > dist[k][j]))
               lowcost[j] = dist[k][j];
       }
   }
}

int main()
{
   int i, j, k, T;
   scanf("%d", &T);
   while(T--)
   {
       scanf("%d%d", &S, &P);
       for(i = 1; i <= P; i++)
           scanf("%lf%lf", &pnt[i].x, &pnt[i].y);

       memset(res, 0, sizeof(res));
       memset(visited, false, sizeof(visited));
       for(i = 1; i < P; i++)
       {
           for(j = i+1; j <= P; j++)
           {
               double xx = pnt[i].x - pnt[j].x;
               double yy = pnt[i].y - pnt[j].y;
               dist[i][j] = sqrt(xx * xx + yy * yy);
               dist[j][i] = dist[i][j];
           }
       }
       Prim();
       sort(res+1, res+P, cmp);
       printf("%.2lf\n", res[S]);
   }
   return 0;
}