POJ-2253-Frogger

链接：https://vjudge.net/problem/POJ-2253

题意：

给n个点，求1->2的路径的每段路中的最大值的最短路。

思路：

刚开始看不懂题。看了题解后看明白的。

直接Dijkstra算法，Dis数组保存到每点路径中的路段的最大值。

代码：

#include <iostream>
#include <stack>
#include <memory.h>
#include <string>
#include <algorithm>
#include <math.h>
#include <iomanip>
using namespace std;
const int MAXN = 200+10;
double Map[MAXN][MAXN];
double Dis[MAXN];
int Vis[MAXN];
int n;

struct Node
{
    double _x;
    double _y;
}node[MAXN];

double get_Len(Node a,Node b)
{
    return sqrt((a._x-b._x)*(a._x-b._x) + (a._y-b._y)*(a._y-b._y));
}

double Dijkstra()
{
    memset(Dis,0, sizeof(Dis));
    memset(Vis,0, sizeof(Vis));
    for (int i = 2;i<=n;i++)
    {
        Dis[i] = Map[1][i];
    }
    Vis[1] = 1;
    for (int i = 1;i<=n;i++)
    {
        int w;
        double small = 9999999.0;
        for (int j = 1; j <= n; j++)
        {
            if (Vis[j] == 0 && small > Dis[j])
            {
                w = j;
                small = Dis[j];
            }
        }

        Vis[w] = 1;

        if (w == 2)
            return small;

        for (int j = 1; j <= n; j++)
        {
            if (Vis[j] == 0)
            {
                //double now = max(small,Map[w][j]);
                //if (now < Dis[j])
                    //Dis[j] = now;
                Dis[j] = min(Dis[j], max(small, Map[w][j]));
            }
        }
    }
}

int main()
{
    int cnt = 0;
    while (cin >> n && n)
    {
        for (int i = 1;i<=n;i++)
        {
            cin >> node[i]._x;
            cin >> node[i]._y;
        }
        for (int i = 1;i<=n;i++)
            for (int j = 1;j<=n;j++)
                Map[i][j] = Map[j][i] = get_Len(node[i],node[j]);

        double v = Dijkstra();
        if (cnt != 0)
            printf("
");
        printf("Scenario #%d
Frog Distance = %.3f
",++cnt,v);
    }
}