糾結(jié)在這句話了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.

題目要求是如果一旦存在一個點不能到達(dá)另一個點就輸出Send Kurdy?

注意處理時跳過邊長超過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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/Commence/p/4014081.html

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