ural 1205. By the Underground or by Foot? Dijkstra

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题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1205

题意描述：给定N个地铁站点坐标，给定步行和地铁速度，求出从A点到B点最小消耗时间；

思路大致就是将距离换算成时间，抽象成单源最短路径问题；

AC代码（这个暴丑，懒得改了）：

//#define _CRT_SECURE_NO_WARNINGS#include <iostream>#include <stdio.h>#include <cstring>#include <string>#include <stack>#include <algorithm>#include <vector>#include <queue>#include <cmath>#include <map>#include <iomanip>using namespace std;vector<vector<double> >v;vector<pair<double, double> >cord;vector<pair<int, double> >node;vector<bool>vis;const int NAL = 99999999;int getOne(){// 找下一个距离最短点double _min = NAL;int idx;for (size_t i = 0; i <vis.size();i++)if (vis[i] == false && _min>node[i].second){_min = node[i].second;idx = i;}return idx;}void func(){double onFoot, onMetro;cin >> onFoot >> onMetro;int N; cin >> N;v.resize(N + 2, vector<double>(N + 2, NAL));cord.resize(N + 2);node.resize(N + 2, make_pair(-1, NAL));vis.resize(N + 2, false);for (int i = 1; i < N + 1; i++){cin >> cord[i].first >> cord[i].second;}int src, tar;while (cin >> src >> tar&&src&&tar){// 算地铁路径时间double tmp = sqrt(pow(cord[src].first - cord[tar].first, 2) + pow(cord[src].second - cord[tar].second, 2));tmp /= onMetro;v[src][tar] = tmp;v[tar][src] = tmp;}cin >> cord[0].first >> cord[0].second;cin >> cord[N + 1].first >> cord[N + 1].second;for (int i = 0; i <= N + 1; i++)for (int j = 0; j <= N + 1; j++){//用步行时间更新状态if (i == j)continue;double tmp = sqrt(pow(cord[i].first - cord[j].first, 2) + pow(cord[i].second - cord[j].second, 2));tmp /= onFoot;if (v[i][j] > tmp)v[i][j] = tmp;}node[0].second = 0;//for (int i = 1; i <= N + 1; i++)//node[i].second = v[0][i];while (true){src = getOne();if (src == N + 1)break;vis[src] = true;for (int i = 0; i <= N + 1; i++)if (i != src&&vis[i] == false && node[i].second > node[src].second + v[src][i]){node[i].second = node[src].second + v[src][i];node[i].first = src;}}cout << setprecision(7) << fixed << node[src].second << endl;stack<int>ans;while (node[src].first != -1){src = node[src].first;ans.push(src);}ans.pop();cout << ans.size();while (!ans.empty()){cout << ' ' << ans.top();ans.pop();}cout << endl;}int main(){//freopen("out.txt", "w", stdout);//freopen("in.txt", "r", stdin);func();}

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