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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.
题目就是求所有通路中，最大边权最小的那一条。。。。。。
应用Dijkstra的思想，一个个的标记。。。
代码如下：
#include<iostream> #include<cstring> #include<cmath> #define max(a,b) (a>b ? a:b) using namespace std; const int INF=10e8; int N; int X[210],Y[210]; double ans[210]; bool vis[210]; void Dijkstra() { int k; double minn,len; for(int i=1;i<=N;++i) { vis[i]=0; ans[i]=INF; } ans[1]=0; for(int i=1;i<=N;++i) { k=-1; minn=INF; for(int j=1;j<=N;++j) if(!vis[j] && ans[j]<minn) { minn=ans[j]; k=j; } if(k==-1) break; vis[k]=1; for(int j=1;j<=N;++j) { len=sqrt((double(X[k])-X[j])*(X[k]-X[j])+(double(Y[k])-Y[j])*(Y[k]-Y[j])); if(!vis[j] && max(len,ans[k])<ans[j]) ans[j]=max(len,ans[k]); } } } int main() { ios::sync_with_stdio(false); cout.setf(ios::fixed); cout.precision(3); int cas=1; for(cin>>N;N;cin>>N,++cas) { for(int i=1;i<=N;++i) cin>>X[i]>>Y[i]; Dijkstra(); cout<<"Scenario #"<<cas<<endl; cout<<"Frog Distance = "<<ans[2]<<endl<<endl; } return 0; }
(简单) POJ 2253 Frogger，Dijkstra。
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原文地址：http://www.cnblogs.com/whywhy/p/4338581.html
