OpenJudge 兔子与樱花

【题解】

求任意两点间的最短路径。此题数据量较小，用Floyd算法，时间复杂度为O(n^3)。

参考https://blog.csdn.net/qq_34594236/article/details/64971883

【代码】

 #include <iostream>
 #include <map>
 #include <string>
 #define maxn 50
 #define INF 100000
 using namespace std;

 int dist[][], path[][];
 map<string, int> map1;
 map<int, string>map2;

 void init()
 {
     int P, Q;
     string scene;
     for (int i = ; i < maxn; i++) {
         for (int j = ; j < maxn; j++) {
             dist[i][j] = INF;
             path[i][j] = j;    // denotes the next point from point i on the shortest path from i to j
         }
         dist[i][i] = ;
     }
     cin >> P;
     for (int i = ; i < P; i++) {
         cin >> scene;
         map1[scene] = i;
         map2[i] = scene;
     }
     cin >> Q;
     for (int i = ; i < Q; i++) {
         string t1, t2;
         int d;
         cin >> t1 >> t2 >> d;
         dist[map1[t1]][map1[t2]] = dist[map1[t2]][map1[t1]] = d;
     }
 }

 void floyd()
 {
     for(int k = ; k < maxn; k++)
         for(int i = ; i < maxn; i++)
             for (int j = ; j < maxn; j++) {
                 if (dist[i][j] > dist[i][k] + dist[k][j]) {
                     dist[i][j] = dist[i][k] + dist[k][j];
                     path[i][j] = path[i][k];
                 }
             }
 }

 int main()
 {
     int R;
     init();
     floyd();
     cin >> R;
     for (int i = ; i < R; i++) {
         string t1, t2;
         int k;
         cin >> t1 >> t2;
         if (t1 == t2) {
             cout << t1 << endl;
             continue;
         }
         k = path[map1[t1]][map1[t2]];
         cout << t1 << "->(" << dist[map1[t1]][k] << ")->";
         while (k != map1[t2]) {
             cout << map2[k] << "->(" << dist[k][path[k][map1[t2]]] << ")->";
             k = path[k][map1[t2]];
         }
         cout << t2 << endl;
     }
     //system("pause");
     return ;
 }