Floyed算法[通俗易懂]

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发布于 2022-09-07 10:46:45

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发布于 2022-09-07 10:46:45

大家好，又见面了，我是你们的朋友全栈君。

这一讲简单介绍一下Floyed算法。话不多说，先放一道题帮助理解（其实是懒得描述具体应用场景）。 Frogger

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

严格来说，这并不是一道标准的Floyed题，题意是：在所有通路中，找到每一条通路中的最大距离，在这所有的距离中，找到一个最小值。之所以也放到这里，首先是因为Floyed算法标志性的三重循环，其次也是想借此拓展一下该算法灵活的应用方式

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define maxsize 1010
#define inf 99999999
int x[maxsize],y[maxsize],sum=1;
double dis[maxsize][maxsize];//用来记录任意两点通路的单步最大长度

//计算两点间的距离
double caldistance(int x1,int y1,int x2,int y2)
{
    return sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1));
}
int main()
{
    int m;
    while(scanf("%d",&m)!=EOF)
    {
        if(m==0)
        {
            break;
        }
        int i,j,k;

        //输入每个点的坐标
        for(i=1;i<=m;i++)
        {
            scanf("%d %d",&x[i],&y[i]);
        }

        //计算任意两个点间的距离
        for(i=1;i<=m;i++)
        {
            for(j=i;j<=m;j++)
            {
                if(j==i) dis[i][j]=0;
                else
                dis[i][j]=dis[j][i]=caldistance(x[i],y[i],x[j],y[j]);
            }
        }
        //算法标志性的三重循环
        //特别暴力的找到所有情况然后更新
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=m;j++)
            {
                for(k=1;k<=m;k++)
                {
                    if(dis[j][k]>max(dis[j][i],dis[i][k]))
                    {
                        dis[j][k]=max(dis[j][i],dis[i][k]);
                        //注意i，j,k的顺序和位置，最初写错了，怎么都找不到问题错在哪里，一定要注意，
                        //如果是找最短路径的正宗Floyed，比较的就是dis[j][k]和dis[j][i]+dis[i][k]的大小，取小的那个
                    }

                }
            }
        }

        printf("Scenario #%d\nFrog Distance = %.3f\n\n", sum++, dis[1][2]);
    }
    return 0;
}

这个题巧妙的一点改动就实现了功能的“巨大”改变。

算法的核心就是那个三重循环及其嵌套顺序。

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