2054 - 青蛙求爱
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.
题目输入
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.
题目输出
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.
输入/输出样例
输入格式
2 0 0 3 4 3 17 4 19 4 18 5 0
输出格式
Scenario #1 Frog Distance = 5.000 Scenario #2 Frog Distance = 1.414
C++解答
//Memory Time //584K 63MS #include<iostream> #include<math.h> #include<iomanip> using namespace std; class coordinate { public: double x,y; }point[201]; double path[201][201]; //两点间的权值 int main(void) { int i,j,k; int cases=1; while(cases) { /*Read in*/ int n; //numbers of stones; cin>>n; if(!n)break; for(i=1;i<=n;i++) cin>>point[i].x>>point[i].y; /*Compute the weights of any two points*/ for(i=1;i<=n-1;i++) for(j=i+1;j<=n;j++) { double x2=point[i].x-point[j].x; double y2=point[i].y-point[j].y; path[i][j]=path[j][i]=sqrt(x2*x2+y2*y2); //双向性 } /*Floyd Algorithm*/ for(k=1;k<=n;k++) //k点是第3点 for(i=1;i<=n-1;i++) //主要针对由i到j的松弛,最终任意两点间的权值都会被分别松弛为最大跳的最小（但每个两点的最小不一定相同） for(j=i+1;j<=n;j++) if(path[i][k]<path[i][j] && path[k][j]<path[i][j]) //当边ik,kj的权值都小于ij时，则走i->k->j路线，否则走i->j路线 if(path[i][k]<path[k][j]) //当走i->k->j路线时，选择max{ik,kj},只有选择最大跳才能保证连通 path[i][j]=path[j][i]=path[k][j]; else path[i][j]=path[j][i]=path[i][k]; cout<<"Scenario #"<<cases++<<endl; cout<<fixed<<setprecision(3)<<"Frog Distance = "<<path[1][2]<<endl; //fixed用固定的小数点位数来显示浮点数（包括小数位全为0) //setprecision(3)设置小数位数为3 cout<<endl; } return 0; }