zoukankan      html  css  js  c++  java
  • 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 == 0 ? 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 = 1;
        scanf("%d",&T);
        while (T--)
        {
            scanf("%d",&N);
            for (int i = 0; i <= N; i++)
               for (int j = 0; j <= N; j++)
                 dp[i][j] = 10000000.0;
            for (int i = 1; i <= N; i++) scanf("%d%d",&x[i],&y[i]);
            for (int i = 1; i <= N; i++)
              for (int j = 1; 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 = 1; k <= N; k++)
              for (int i = 1; i <= N; i++)
                for (int j = 1; j <= N; j++)
                  dp[i][j] = min(dp[i][j],dp[i][k] + dp[k][j]);
            double ans = 0.0;
            for (int i = 1; i <= N; i++)
               for (int j = 1; j <= N; j++)
            {
               ans = max(ans,dp[i][j]);
            }
            printf("Case #%d:
    ",kase++);
            if (ans == 10000000.0) puts("Send Kurdy");
            else printf("%.4lf
    ",ans);
            putchar('
    ');
        }
        return 0;
    }
  • 相关阅读:
    面试常考点:http和https的区别与联系
    常见的反爬虫和应对方法
    2019/1/1 Python今日收获
    2018/12/26,12/27 Python今日收获
    2018/6/7-6/8 Python今日收获
    2018/6/6 Python今日收获
    CSS(3)——visited伪类
    CSS中margin和padding的区别
    CSS(2)——CSS的文字,边框,背景与列表
    CSS(1)——CSS的引入方式与选择器
  • 原文地址:https://www.cnblogs.com/Commence/p/4014081.html
Copyright © 2011-2022 走看看