POJ 2253 Frogger(Floyd Dij Spfa變形)

大白QQly成長日記發表於2018-09-06

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

題目大意：求2點間所有路中最小的最大距離。就是每條路有一個最大2點間距離，求所有路中最小的。

思路：資料小，弗洛伊德，迪杰特斯拉，SPFA都能做，暑假這些忘了不少，拾遺吧。

弗洛伊德：3重迴圈，暴力求所有點之間的最短路 時間複雜度n^3

迪杰特斯拉：以起點為原點不斷擴張找最小，以鄰點為媒介找更小縮排。  時間複雜度n^2

SPFA：通過佇列進行優化，

Floyd程式碼：

#include<set>
#include<map>
#include<list>
#include<deque>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<stdio.h>
#include<sstream>
#include<stdlib.h>
#include<string.h>
//#include<ext/rope>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define per(i,a,b) for(int i=a;i<=b;++i)
#define LL long long 
#define swap(a,b) {int t=a;a=b;b=t} 
using namespace std;
//using namespace __gnu_cxx;
int a[300],b[300];
double p[305][305];
int main()
{
	int t,z=0;
	while(scanf("%d",&t)!=EOF&&t!=0)
	{
		z++;
		memset(p,0,sizeof(p));
		per(i,1,t) scanf("%d %d",&a[i],&b[i]);
		per(i,1,t)
		{
			per(j,i+1,t)
			{
				p[j][i]=p[i][j]=sqrt(double(a[i]-a[j])*(a[i]-a[j])+double(b[i]-b[j])*(b[i]-b[j]));
			}
		}
		per(k,1,t)
		{
			per(i,1,t)
			{
				per(j,1,t)
				{
				    p[i][j]=min(p[i][j],max(p[i][k],p[k][j]));
				}
			}
		} 
		printf("Scenario #%d\n",z);
		printf("Frog Distance = %.3lf\n\n",p[1][2]);
	}
	return 0; 
}

Dij程式碼：

#include<set>
#include<map>
#include<list>
#include<deque>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<stdio.h>
#include<sstream>
#include<stdlib.h>
#include<string.h>
//#include<ext/rope>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define per(i,a,b) for(int i=a;i<=b;++i)
#define LL long long 
#define swap(a,b) {int t=a;a=b;b=t} 
using namespace std;
//using namespace __gnu_cxx;
int n;
struct node{
	int x;
	int y;
}s[300];
double p[300][300];
int vis[300];
double d[300];
void dij(int z)
{
    fill(d,d+300,INF);
    d[z]=0;
    per(i,1,n) 
    {
    	int k,minn=INF;
		per(j,1,n) 
		{
			if(vis[j]==0&&d[j]<minn)//臨接點距離最小的 
			{
				k=j;
				minn=d[j];
			}
		}
		vis[k]=1;
		per(j,1,n)// 
		{
			/*if(vis[j]==0&&p[k][j]!=INF&&d[k]+p[k][j]<d[j])//以k為起點，可以到達j且小於j到原點的距離。縮排 
			{
				d[j]=d[k]+p[k][j];
			}*/
			d[j]=min((double)d[j],max((double)d[k],p[k][j]));
		}
	}
}
int main()
{
    int k=1;
    while(~scanf("%d",&n)&&n!=0)
    {
        memset(p,0,sizeof(p));
        memset(vis,0,sizeof(vis)); 
        per(i,1,n) scanf("%d%d",&s[i].x,&s[i].y);
        per(i,1,n)
            per(j,i+1,n)
            {
				p[i][j]=p[j][i]=sqrt(double(s[i].x-s[j].x)*(s[i].x-s[j].x)+double(s[i].y-s[j].y)*(s[i].y-s[j].y));
			}
        dij(1);
        //per(i,1,n) cout<<d[i]<<"*";
        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",k++,d[2]);
    }
    return 0;
}

SPFA：

#include<set>
#include<map>
#include<list>
#include<deque>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<stdio.h>
#include<sstream>
#include<stdlib.h>
#include<string.h>
//#include<ext/rope>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define per(i,a,b) for(int i=a;i<=b;++i)
#define LL long long 
#define swap(a,b) {int t=a;a=b;b=t} 
using namespace std;
//using namespace __gnu_cxx;
int n;
struct node{
	int x;
	int y;
}s[300];
double p[300][300];
int vis[300];
double d[300];
/*void dij(int z)
{
    fill(d,d+300,INF);
    memset(vis,0,sizeof(vis));
    d[z]=0;
    per(i,1,n)
    {
    	int minn=INF,k;
    	per(j,1,n)
    	{
    		if(vis[j]==0&&d[j]<minn)
    		{
    			k=j;
    			minn=d[j];
			}
		}
		vis[k]=1;
		per(j,1,n)
		{
			/*if(vis[j]==0&&p[k][j]!=INF&&d[k]+p[k][j]<d[j])
			{
				d[j]=d[k]+p[k][j];
			}
			d[j]=min(d[j],max(d[k],p[j][k]));
		}
	}
}*/
void spfa()
{
    queue<int>q;
    fill(d,d+300,INF);
    d[1]=0;
    for(int i=1; i<=n; i++)
        vis[i]=0;
    vis[1]=1;
    q.push(1);
    while(!q.empty())
    {
        int k=q.front();
        vis[k]=0;
        q.pop();
        for(int j=1; j<=n; j++)
            if(max(d[k],p[k][j])<d[j])
            {
                d[j]=max(d[k],p[k][j]);
                if(vis[j]==0)
                {
                    q.push(j);
                    vis[j]=1;
                }
            }
    }
}
int main()
{
    int k=1;
    while(~scanf("%d",&n)&&n!=0)
    {
        memset(p,0,sizeof(p));
        memset(vis,0,sizeof(vis)); 
        per(i,1,n) scanf("%d%d",&s[i].x,&s[i].y);
        per(i,1,n)
            per(j,i+1,n)
            {
				p[i][j]=p[j][i]=sqrt(double(s[i].x-s[j].x)*(s[i].x-s[j].x)+double(s[i].y-s[j].y)*(s[i].y-s[j].y));
			}
        spfa();
        //per(i,1,n) cout<<d[i]<<"*";
        printf("Scenario #%d\nFrog Distance = %.3lf\n\n",k++,d[2]);
    }
    return 0;
}

 

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