Thunder Mountain - UVa 10803 Floyd

最新推荐文章于 2021-02-27 22:53:47 发布

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面对危险的野外环境，本文介绍了一种算法解决方案，用于确保在不同城镇间安全有效地运输物资，即使在恶劣条件下也能找到最远不超过10公里的路径。

Problem ?
Thunder Mountain
Time Limit: 3 seconds

"I mean, some people got guns, and some
people got flashlights, and some people got
batteries. These guys had all three."

J. Michael Straczynski, "Jeremiah."

Markus is building an army to fight the evil Valhalla Sector, so he needs to move some supplies between several of the nearby towns. The woods are full of robbers and other unfriendly folk, so it's dangerous to travel far. As Thunder Mountain's head of security, Lee thinks that it is unsafe to carry supplies for more than 10km without visiting a town. Markus wants to know how far one would need to travel to get from one town to another in the worst case.

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n (the number of towns, 1<n<101). The next n lines will give the xy-locations of each town in km (integers in the range [0, 1023]). Assume that the Earth is flat and the whole 1024x1024 grid is covered by a forest with roads connecting each pair of towns that are no further than 10km away from each other.

Output
For each test case, output the line "Case #x:", where x is the number of the test case. On the next line, print the maximum distance one has to travel from town A to town B (for some A and B). Round the answer to 4 decimal places. Every answer will obey the formula

fabs(ans*1e4 - floor(ans*1e4) - 0.5) > 1e-2

If it is impossible to get from some town to some other town, print "Send Kurdy" instead. Put an empty line after each test case.

Sample Input	Sample Output
2 5 0 0 10 0 10 10 13 10 13 14 2 0 0 10 1	Case #1: 25.0000 Case #2: Send Kurdy

题意：给你n个点的坐标，问是否每两个点之间都可以通过点与点不超过10的路径连接起来。如果可以的话，输出最大的需要路线距离。

思路：简单的Floyd，没什么好说的。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
double INF=1e9,dis[110][110],x[110],y[110],eps=1e-8;
int T,t,n;
void Floyd()
{
    int i,j,k;
    for(k=1;k<=n;k++)
       for(i=1;i<=n;i++)
          for(j=1;j<=n;j++)
             dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}
int main()
{
    int i,j,k;
    double ret;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
           scanf("%lf%lf",&x[i],&y[i]);
        for(i=1;i<=n;i++)
           for(j=i;j<=n;j++)
           {
               ret=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
               if(ret<10+eps)
                dis[i][j]=dis[j][i]=ret;
               else
                 dis[i][j]=dis[j][i]=INF;
           }
        Floyd();
        printf("Case #%d:\n",t);
        ret=0;
        for(i=1;i<=n;i++)
           for(j=1;j<=n;j++)
              ret=max(ret,dis[i][j]);
        if(ret<INF-eps)
          printf("%.4f\n\n",ret);
        else
          printf("Send Kurdy\n\n");
    }
}