poj 2253 Frogger-优快云博客

本文链接：https://blog.youkuaiyun.com/Code_KK/article/details/47279271

本文通过两种算法——Dijkstra算法和Floyd算法解决了一个有趣的几何问题：如何计算两块石头之间的最小跳跃距离，使得一只青蛙能够从一块石头跳到另一块石头。此问题涉及到计算不同路径上的最大跳跃长度，并找到所有可能路径中的最小值。

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不要忘了最后的%lf改成%f，26遍WA就败在这上面了

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Dijkstra算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
#define MAX 210
using namespace std;

struct node
{
    int x,y;
} E[MAX];

bool vis[MAX];
double dis[MAX];
int n;

double D(int x1,int y1,int x2,int y2)
{
    return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
}
double Dijkstra()
{
    memset(vis,false,sizeof(vis));
    for(int i=1; i<=n; i++)
        dis[i] = INF;
    int pos = 1;
    dis[1] = 0;
    vis[1] = true;
    double ans=INF;
    for(int i=1; i<=n; i++)
    {
        double M = INF;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] < M)
            {
                pos = j;
                M = dis[j];
            }
        }
        if(pos == 2)
            break;
        vis[pos] = true;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && dis[j] > max(dis[pos],D(E[pos].x,E[pos].y,E[j].x,E[j].y)))
                dis[j] = max(dis[pos],D(E[pos].x,E[pos].y,E[j].x,E[j].y));
        }
    }
    return dis[2];
}
int main()
{
    int t = 0;
    while(~scanf("%d",&n) && n)
    {
        for(int i=1; i<=n; i++)
            scanf("%d%d",&E[i].x,&E[i].y);
        double ant = Dijkstra();
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",++t,ant);
    }
    return 0;
}

Floyd 算法

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

#define INF 0x3f3f3f3f

using namespace std;

struct node
{
    int x,y;
} ls[220];

double dis[220][220];

int n;

double Floyd()
{
    for(int k=1; k<=n; k++)
    {
        for(int i=1; i<n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                if(dis[i][k] < dis[i][j] && dis[k][j] < dis[i][j])
                {
                    if(dis[i][k] < dis[k][j])
                    {
                        dis[i][j] = dis[j][i] = dis[k][j];
                    }
                    else
                    {
                        dis[i][j] = dis[j][i] = dis[i][k];
                    }
                }
            }
        }
    }
    return dis[1][2];
}

int main()
{
    //freopen("in.txt","r",stdin);
    int t = 0;
    while(~scanf("%d",&n) && n)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(i!=j)
                    dis[i][j] = INF;
                else
                    dis[i][j] = 0;
            }
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d %d",&ls[i].x,&ls[i].y);
        }
        for(int i=1; i<n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                dis[i][j] = dis[j][i] = sqrt((ls[i].x-ls[j].x)*(ls[i].x-ls[j].x) + (ls[i].y-ls[j].y)*(ls[i].y-ls[j].y));
            }
        }
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",++t,Floyd());
    }

    return 0;
}