poj 2253 最短路 or 最小生成树

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

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

题意：要求青蛙从第一个点跳到第二个点，其他的点作为跳板，可以选择跳或者不跳。问最小权值中的最大值是多少。

注意：这题wA 了好久，因为精度的问题，c++交可以过，g++就wa了

　　　还有 sqrt(x), 里面的 x得是浮点型的数，不然交的时候给你提示编译错误

代码：

const int inf = 1<<29;

struct qnode

{

    double x, y;

}arr[205];

int n;

double fun(int i, int j){

    double ff = sqrt((arr[i].x - arr[j].x)*(arr[i].x - arr[j].x) + (arr[i].y - arr[j].y)*(arr[i].y - arr[j].y));

    return ff;

}

double edge[205][205];

double ans;

struct node

{

    int v;

    double c;

    node(int _v, double _c):v(_v), c(_c){}

    friend bool operator< (node n1, node n2){

        return n1.c > n2.c;

    }

};

double d[205];

bool vis[205];

void prim(){

    ans = 0;

    priority_queue<node>que;

    while(!que.empty()) que.pop();

    for(int i = 1; i <= n; i++){

        d[i] = edge[1][i];

        que.push(node(i, d[i]));

    }

    memset(vis, false, sizeof(vis));

    while(!que.empty()){

        node tem = que.top();

        que.pop();

        int v = tem.v;

        double c = tem.c;

        ans = max(ans, c);

        if (v == 2) return;

        if (vis[v]) continue;

        vis[v] = true;;

        for(int i = 1; i <= n; i++){

            if (!vis[i] && edge[v][i] < d[i]){

                d[i] = edge[v][i];

                que.push(node(i, d[i]));

            }

        }

    }

}

int main() {

    int k = 1;

    while(~scanf("%d", &n) && n){

        for(int i = 1; i <= n; i++){

            scanf("%lf%lf", &arr[i].x, &arr[i].y);

        }

        for(int i = 1; i <= n; i++){

            for(int j = 1; j <= n; j++)

                edge[i][j] = inf;

        }

        for(int i = 1; i <= n; i++){

            for(int j = i+1; j <= n; j++){

                double len = fun(i, j);

                edge[i][j] = edge[j][i] = len;

            }

        }

        prim();

        printf("Scenario #%d\n", k++);

        printf("Frog Distance = %.3f\n\n", ans);

    }

    return 0;

}

巴特西

poj 2253 最短路 or 最小生成树

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