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[50][50], path[50][50];
map<string, int> map1;
map<int, string>map2;

void init()
{
    int P, Q;
    string scene;
    for (int i = 0; i < maxn; i++) {
        for (int j = 0; 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] = 0;
    }
    cin >> P;
    for (int i = 0; i < P; i++) {
        cin >> scene;
        map1[scene] = i;
        map2[i] = scene;
    }
    cin >> Q;
    for (int i = 0; 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 = 0; k < maxn; k++)
        for(int i = 0; i < maxn; i++)
            for (int j = 0; 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 = 0; 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 0;
}