Poj 2253 Frogger(floyd||SPFA的变形)

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28785		Accepted: 9348

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

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

求最小的最大距离。

//用G++WA 而c++AC。。。- -。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<math.h>
using namespace std;
double dis[300][300];
int n,m,s=0;
struct ndoe
{
    int  a,b;
}q[300];
double max(double x,double y)
{
    return x>y?x:y;
}
double min(double x,double y)
{
    return x<y?x:y;
}
void floyd()
{
    int i,j,k;
    for(k=0;k<n;k++)
    {
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                dis[i][j]=min( dis[i][j],max(dis[i][k],dis[k][j]) );//求最小的最大距离，外没有if(dis[i][j]>dis[i][k]+dis[k][j])条件
            }
        }
    }
}
int main()
{
    int i,j,k,x,y;
    while(cin>>n&&n)
    {
        s++;
        for(i=0;i<n;i++)
        {
            cin>>q[i].a>>q[i].b;//注意对于点的坐标形式，要用结构体或数组来装。
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(i!=j)
                dis[i][j]=sqrt( (q[i].a-q[j].a)*1.0*(q[i].a-q[j].a)+(q[i].b-q[j].b)*1.0*(q[i].b-q[j].b)  );
                else dis[i][j]=0;
            }
        }
      floyd();
      printf("Scenario #%d\n",s);
      printf("Frog Distance = %.3lf\n",dis[0][1]);
      printf("\n");
    }
    return 0;
}

法二：SPFA求解最短路中的最小中的最大路径长度。（PS：POJ中最好不要用lf而用f）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<queue>
#include<cmath>
#include<algorithm>
#define MAX 1000
#define inf 0x3f3f3f3f
using namespace std;
struct node{
    double a,b;
}q[300];
double a[400][400],dis[400];
bool vis[4000];
int n;
void SPFA(){
    for(int i=0;i<=n;i++){
       vis[i]=false;
    }
    fill(dis,dis+n,inf);///注意不要对浮点类型的数组进行memset()的初始化，要用循环或者是fill
    memset(vis,false,sizeof(vis));
    queue<int>q;
    while(!q.empty())
        q.pop();
    dis[0]=0;
    q.push(0);
    vis[0]=true;

    while(!q.empty()){

        int v=q.front();
        q.pop();vis[v]=false;
        for(int i=0;i<n;i++){
            if(dis[i]>max(dis[v],a[v][i]) ){
                dis[i]=max(dis[v],a[v][i]);
                if(!vis[i]){
                    vis[i]=true;
                    q.push(i);
                }
            }
        }
    }
}
int main(){
    int m,i,j,k,cnt=1;
    while(~scanf("%d",&n)&n){
        for(i=0;i<n;i++){
            scanf("%lf%lf",&q[i].a,&q[i].b);
        }
        for(i=0;i<n;i++){
            for(j=0;j<=i;j++){
                a[i][j]=a[j][i]=sqrt((q[i].a-q[j].a)*(q[i].a-q[j].a)+(q[i].b-q[j].b)*(q[i].b-q[j].b) );
                //printf("%.2lf\n",a[i][j]);
            }
        }
        SPFA();
        printf("Scenario #%d\n",cnt++);
        printf("Frog Distance = %.3f\n\n",dis[1]);
    }
    return 0;
}