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原题地址：http://poj.org/problem?id=2502
Subway
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Time Limit: 1000MS |
Memory Limit: 65536K |
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Total Submissions: 7347 |
Accepted: 2387 |
Description
You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don‘t want to be late for class, you want to know how long it will take you to get to school.
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you
are lucky, and whenever you arrive at a subway station, a train is there that
you can board immediately. You may get on and off the subway any number of
times, and you may switch between different subway lines if you wish. All
subway lines go in both directions.
Input
Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.
Output
Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.
Sample Input
0 0 10000 1000
0 200 5000 200 7000 200 -1 -1
2000 600 5000 600 10000 600 -1 -1
Sample Output
21
题意：
只有一组数据，第一行给的是家的坐标和学校的坐标，接着后面的都是地铁线路上各个站的坐标，以(-1,-1)结束。已知步行速度为10km/h，地铁速度为40km/h，求家里到学校的最短时间（分钟，四舍五入到整数）。
注意：
#include <cstdio> #include <math.h> #include <algorithm> using namespace std; const int N = 1000, INF = 0x3f3f3f3f; struct POINT { int x; int y; }vertex[N]; double time[N][N] = {0}; bool cnt[N] = {0}; void qwe(int& n) { int x, y, start, end; while(1) { start = n; for(; ; ++n) { if(!~scanf("%d %d", &x, &y)) return ; if(x == -1 && y == -1) { end = n; break; } else { vertex[n].x = x; vertex[n].y = y; } } for(int i=start+1; i<end; ++i) { double dx = (vertex[i].x - vertex[i-1].x) / 1000.0; double dy = (vertex[i].y - vertex[i-1].y) / 1000.0; time[i][i-1] = sqrt(dx*dx + dy*dy) * 1.5; time[i-1][i] = time[i][i-1]; } } } int main(void) { scanf("%d %d %d %d", &vertex[0].x, &vertex[0].y, &vertex[1].x, &vertex[1].y); int n = 2, mintag; qwe(n); for(int i=0; i<n; ++i) { time[i][i] = INF; for(int j=0; j<n; ++j) { if(time[i][j] == 0) { double dx = (vertex[i].x - vertex[j].x) / 1000.0; double dy = (vertex[i].y - vertex[j].y) / 1000.0; time[i][j] = sqrt(dx*dx + dy*dy) * 6; time[j][i] = time[i][j]; } } } cnt[0] = true; for(int i=1; i<n; ++i) { mintag = 0; for(int j=1; j<n; ++j) if(!cnt[j] && time[0][j] < time[0][mintag]) mintag = j; cnt[mintag] = true; for(int j=1; j<n; ++j) if(!cnt[j]) time[0][j] = min(time[0][j], time[0][mintag] + time[mintag][j]); } printf("%.0lf\n", time[0][1]); }
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原文地址：http://www.cnblogs.com/gwtan/p/4780202.html
