UVA 10803 Thunder Mountain

纠结在这句话了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.

题目要求是如果一旦存在一个点不能到达另一个点就输出Send Kurdy

注意处理时跳过边长超过10的再跑FLOYD。之后在所有最短路中查找最大值即可

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b ==  ? b : gcd(b, a % b);}
#define MAXN 105
double dp[MAXN][MAXN];
int x[MAXN],y[MAXN];
int N;
int main()
{
    //freopen("sample.txt","r",stdin);
    int T,kase = ;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&N);
        for (int i = ; i <= N; i++)
           for (int j = ; j <= N; j++)
             dp[i][j] = 10000000.0;
        for (int i = ; i <= N; i++) scanf("%d%d",&x[i],&y[i]);
        for (int i = ; i <= N; i++)
          for (int j = ; j <= N; j++)
        {
          double tmp = (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) *(y[i] - y[j]);
          if (tmp > 100.0) continue;
          dp[i][j] = min(dp[i][j],sqrt(tmp));
        }
        for (int k = ; k <= N; k++)
          for (int i = ; i <= N; i++)
            for (int j = ; j <= N; j++)
              dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j]);
        double ans = 0.0;
        for (int i = ; i <= N; i++)
           for (int j = ; j <= N; j++)
        {
           ans = max(ans,dp[i][j]);
        }
        printf("Case #%d:\n",kase++);
        if (ans == 10000000.0) puts("Send Kurdy");
        else printf("%.4lf\n",ans);
        putchar('\n');
    }
    return ;
}