POJ 2253 Frogger

解题思路:最短路径dijkstra算法

#include<iostream>
#include <cmath>
#include <iomanip>
using namespace std;

#define MaxNum 1e8 
int main()
{
    int n, start, tempS;
    bool visited[205];
    double dist[205];
    double  minDist, tempDist;
    int cases = 0;
    struct  
    {
        int x, y;
    }p[205];

    while (cin >> n && n)
    {
        cases++;
        memset(visited, 0, sizeof(bool) * 205);
        fill(&dist[0], &dist[204], MaxNum);

        for (int i = 0; i < n; i++)
            cin >> p[i].x >> p[i].y;

        visited[0] = true;
        dist[0] = 0;
        start = 0;

        for (int i = 0; i < n; i++)
        {
            minDist = MaxNum;
            for (int j = 0; j < n; j++)
            {
                if (!visited[j])
                {
                    tempDist = pow(double(p[start].x - p[j].x), 2);
                    tempDist += pow(double(p[start].y - p[j].y), 2);
                    tempDist = sqrt(tempDist);
                    tempDist = max(tempDist, dist[start]);
                    dist[j] = min(dist[j], tempDist);

                    if (minDist > dist[j])
                    {
                        minDist = dist[j];
                        tempS = j;
                    }
                }
            }
            start = tempS;
            visited[start] = true;
            if (start == 1)
                break;
        }
        if (cases != 1)
            cout << "\n";
        cout << "Scenario #" << cases << endl;
        cout << "Frog Distance = " << fixed << setprecision(3) << dist[1] << endl;
    }
    return 0;
}