POJ2253《Frogger》方法：Floyd-Wallshall

来源：互联网发布：判断矩阵权重计算编辑：程序博客网时间：2024/05/29 18:24

将Floyd算法中求每对顶点间的最小距离，条件换成每对顶点间最长序列的最大跳数，所以要多加进去k。

#include <iostream>#include <cstdio>#include <cmath>#include <iomanip>using namespace std;class coordinate {public:double x, y;} points[201];double path[201][201];int main(){//freopen("temp.txt", "r", stdin);int n;int cnt = 1;while (cin >> n) {if (n == 0)break;for (int i = 1; i <= n; ++i) cin >> points[i].x >> points[i].y;for (int i = 1; i <= n-1; ++i) {for (int j = i+1; j <= n; ++j) {double x = points[i].x - points[j].x;double y = points[i].y - points[j].y;path[i][j] = path[j][i] = sqrt(x*x + y*y); //双向性}}// Floyd-Wallshall// 这里将原本的求两点间最短路径，变为求两点间路径的最大跳数for (int k = 1; k <= n; ++k) {for (int i = 1; i <= n-1; ++i) {for (int j = i+1; j <= n; ++j) {if (path[i][k] < path[i][j] && path[k][j] < path[i][j]) {if (path[i][k] < path[k][j]) {path[i][j] = path[j][i] = path[k][j]; //不要忘了双向性} else {path[i][j] = path[j][i] = path[i][k];}}}}}cout << "Scenario #" << cnt++ << endl;cout << "Frog Distance = " << fixed << setprecision(3) << path[1][2] << endl << endl;}return 0;}