poj 2253 Frogger 【枚举+并查集 or 最短路】

Frogger

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28514		Accepted: 9263

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

题意：给出n个点的坐标，连通第一个点和第二个点有很多条路径，每条路径都会有 一个最大距离d - ->（当前路径经过的点集里面 任意两点间的距离都小于或者等于d）。题目要求找到这样的一个路径：（1）连通1，2两点；（2）这条路径中的最大距离d 是所有可选择路径中最小的。  输出这条路径的最大距离d。

枚举+并查集：16ms （还是这个思路好想） 按权值升序排列，从前到后枚举并判断1，2是否连通即可。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f
#define MAX 200+10 
using namespace std;
int n;
int k = 1;
int m;
int set[MAX];
double x[MAX], y[MAX];
struct rec
{
    int s, e;
    double d;
}num[200*200+10];
bool cmp(rec a, rec b)
{
    return a.d < b.d;
}
double dis(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void init()
{
    for(int i = 1; i <= n; i++)
    set[i] = i;
}
int find(int p)
{
    int t;
    int child = p;
    while(p != set[p])
    p =set[p];
    while(child != p)
    {
        t = set[child];
        set[child] = p;
        child = t;
    }
    return p;
}
void merge(int x, int y)
{
    int fx = find(x);
    int fy = find(y);
    if(fx != fy)
    set[fx] = fy;
}
void getmap()
{
    int i, j;
    m = 0;
    for(i = 1; i <= n; i++)
    scanf("%lf%lf", &x[i], &y[i]);
    for(i = 1; i <= n; i++)
    {
        for(j = i+1; j <= n; j++)
        {
            num[m].s = i;
            num[m].e = j;
            num[m++].d = dis(x[i], y[i], x[j], y[j]);
        }
    }
    sort(num, num+m, cmp);
}
void slove()
{
    int i, j;
    int exist;
    double ans;
    for(i = 0; i < m; i++)
    {
        init();//初始化 
        ans = num[i].d;
        exist = 0;
        for(j = i; j < m; j++)
        {
            ans = num[j].d;//选择最大的 
            merge(num[j].s, num[j].e);
            if(find(1) == find(2))
            {
                exist = 1;//连通
                break; 
            }
        }
        if(exist)//第一个满足的一定是最小的 
        {
            printf("Scenario #%d\nFrog Distance = %.3lf\n\n", k++, ans);
            return ;
        }
    }
}
int main()
{
    while(scanf("%d", &n), n)
    {
        getmap();//建图 
        slove();
    }
    return 0;
}

最短路径变形:16ms 错了两次 醉了。。。

思路：dijkstra算法 更新环节dist[j] = min(dist[j], dist[next] + map[next][j])（更新最短距离，一个距离是源点到该点的！当前！最短距离 ，另一个是通过媒介点所得到的距离）。                                                                                                          按照题目要求，两个过程各自对应一个最大距离d，而我们要的是这两个最大距离里面的较小的值，即dist[j] = min(max(dist[next], map[next][j]),dist[j]).

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f
#define MAX 200+10
using namespace std;
double map[MAX][MAX], dist[MAX];
double x[MAX], y[MAX];
int vis[MAX];
int n, p = 1;
double dis(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
void init()
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            if(i == j)
            map[i][j] = 0;
            else
            map[i][j] = INF;
        }
    }
}
void getmap()
{
    int i, j;
    for(i = 1; i <= n; i++)
    scanf("%lf%lf", &x[i], &y[i]);
    for(i = 1; i <= n; i++)
    {
        for(j = i+1; j <= n; j++)
        {
            map[i][j] = map[j][i] = dis(x[i], y[i], x[j], y[j]);
        }
    }
}
void dijkstra()
{
    int i, j, next;
    double Min;
    int t = 0;
    for(i = 1; i <= n; i++)
    {
        vis[i] = 0;
        dist[i] = map[1][i];
    }
    vis[1] = 1;
    for(i = 2; i <= n; i++)
    {
        Min = INF;
        for(j = 1; j <= n; j++)
        {
            if(!vis[j] && Min > dist[j])
            {
                Min = dist[j];
                next = j;
            }
        }
        vis[next] = 1;
        for(j = 1; j <= n; j++)
        {
            if(!vis[j])
            {
                dist[j] = min(max(map[next][j], dist[next]), dist[j]);
            }  
        }
    }
    printf("Scenario #%d\nFrog Distance = %.3lf\n\n", p++, dist[2]);
}
int main()
{
    while(scanf("%d", &n), n)
    {
        init();//初始化 
        getmap();//建图 
        dijkstra();
    } 
    return 0;
}