poj2253 变形Floyd

//poj2253 求青蛙最小的一跳的距离。同时又可以到达目的地。//构造一个全部点的距离相邻矩阵 //然后使用变形的Floyd 如果存在 AB < AC BC < AC, 则AC == max(AB, BC) //前两次TLE...MD  后面发现原来是多加了一层松弛，去掉，然后对换遍历的顺序就可以了。 #include <iostream>#include <cmath>#include <iomanip>#include <algorithm>using namespace std;double dis[205][205]; double x[205];double y[205];double max(double a, double b){return a < b? b:a;}void Dij(int n){for(int k = 0; k < n; k++){for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){if(dis[i][j] > dis[i][k] && dis[i][j] > dis[k][j]){dis[i][j] = max(dis[i][k], dis[k][j]);}}}}}int main(){int n;cin>>n;int x1, y1, x2, y3, tempx, tempy;int k = 1;while(n){memset(dis, 0, sizeof(dis));memset(x, 0, sizeof(x));memset(y, 0, sizeof(y));for(int i = 0; i < n; i++){cin>>x[i]>>y[i];}for(int i = 0; i < n; i++){for(int j = i + 1; j < n; j++){dis[i][j] = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);dis[j][i] = dis[i][j];}}Dij(n);cout<<"Scenario #"<<k++<<endl;cout<<"Frog Distance = "<<fixed<<setprecision(3)<<sqrt(dis[0][1])<<endl<<endl;cin>>n;}return 0;}