图论_最短路_例题_Frogger(POJ 2253)

Frogger(POJ 2253)

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【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.

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【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.

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【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.

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【Sample Input】

2
0 0
3 4

3
17 4
19 4
18 5

0

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Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

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【解析】

模板题，与最短路的思想一致，得出所有边，并将边排序。从最短的边开始，不断的尝试加入，若该边的两点不在一个集合中，则加入该边；当起始点和终点在同一个集合中时，输出最后加入的边的权值。

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【代码】

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;struct edge{    int v, u;    float w;    bool operator  < (const edge x) const    {        if(w<x.w || (x.w==x.w && (v<x.v || (v==x.v && u<x.u))))            return true;        else return false;    }};pair<int ,int> p[256];edge e[400000];int fa[256];int findfa(int x){    if(fa[x]==-1 || fa[x]==x) return fa[x] = x;    else return fa[x] = findfa(fa[x]);}int main(){    int n;    int CASE = 0;    while(scanf("%d", &n)==1 && n)    {        int m = 0;        memset(fa, -1, sizeof(fa));        scanf("%d%d", &p[1].first, &p[1].second);        scanf("%d%d", &p[n].first, &p[n].second);        for(int i=2; i<n; i++)            scanf("%d%d", &p[i].first, &p[i].second);        for(int i=1; i<n; i++)            for(int j=i+1; j<=n; j++)            {                float x0 = float(p[i].first-p[j].first);                float y0 = float(p[i].second-p[j].second);                e[m++] = (edge){i,j,sqrt(x0*x0+y0*y0)};            }        sort(e, e+m);        int index;        for(int i=0; i<m; i++)        {            findfa(e[i].v);            findfa(e[i].u);            if(fa[e[i].u]==fa[e[i].v]) continue;            fa[fa[e[i].u]] = fa[e[i].v];            if(findfa(1)==findfa(n))            {                printf("Scenario #%d\nFrog Distance = %.3f\n\n", ++CASE, e[i].w);                break;            }        }    }    return 0;}