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最新推荐文章于 2017-07-16 13:24:00 发布

转载最新推荐文章于 2017-07-16 13:24:00 发布 · 131 阅读

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原文链接：http://www.cnblogs.com/Commence/p/4014081.html

本文详细解析了一道编程题，使用Floyd算法解决路径寻址问题，特别关注当无法从一个城镇到达另一个城镇时输出'SendKurdy'的情况。通过跳过边长超过10的路径优化算法，最终找到所有最短路径中的最大值。

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:\n",kase++);
        if (ans == 10000000.0) puts("Send Kurdy");
        else printf("%.4lf\n",ans);
        putchar('\n');
    }
    return 0;
}

转载于:https://www.cnblogs.com/Commence/p/4014081.html