sta col 解題思路 最短 節點 eal rop esc memset

題目連接：http://poj.org/problem?id=2253

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. 題目大意： 青蛙A想到B那裏去，他通過跳石頭的方式過去，給出A，B所在位置和石頭位置的坐標，要求得出所有路徑中最大距離的最小值（A要到B可以選擇不同路徑，每條路徑跳躍若幹石頭，每條路徑都會有一個最大跳躍距離，即該路徑中最遠的兩塊石頭的距離，要求求出這些路徑中最大跳躍距離的最小值） 解題思路： 最短路徑變形，原來求單源最短路徑算法中dis數組用來儲存起點到該節點距離，現在可以用來儲存到達該點的路徑中最大距離的最小值，相應的松弛操作也要修改。 代碼如下：

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

#define MAX 205
#define INF 0x3ffffff
using namespace std;

int n;
double x[MAX];
double y[MAX];
double d[MAX];
bool vis[MAX];

void init(void)
{
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n;i++)
        d[i]=INF;
}

void dijkstra(void)
{
    d[0]=0;
    for(int i=0;i<n;i++)
    {
        int now,m=INF;
        for(int j=0;j<n;j++)
        {
            if(!vis[j]&&d[j]<m)
            {
                now=j;
                m=d[j];
            }
        }
        vis[now]=1;
        for(int j=0;j<n;j++)
        {
            int xx=abs(x[now]-x[j]);
            int yy=abs(y[now]-y[j]);
            double dis=sqrt(pow(xx,2.0)+pow(yy,2.0));
            d[j]=min(d[j],max(d[now],dis));
        }
    }
}

int main()
{
    int T=0;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            return 0;
        T++;
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&x[i],&y[i]);
        init();
        dijkstra();
        cout<<"Scenario #"<<T<<endl;
        cout<<"Frog Distance = ";
        printf("%.3lf\n",d[1]);
        cout<<endl;
    }
}

思考：對於類似的題目，如果是能夠在節點之間轉移的屬性，應該都可以通過類似的方法求出來。

poj2253 最短路變形