华为oj 字串的连接最长路径查找

这道题应该是初级中最难的了吧，这道题整体思路应该是，把每个字符串看成一个节点，这样我们要求的就是在一个有向图中两点形成的最长路径，对于这种类型的题目，可以考虑采用佛洛依德算法，因为它是查找有向图所有两点之间的路径长度，这样很容易就会找到最长的路径！！！

程序如下：

//弗洛伊德算法求最短路径，动态规划,有待深究//dp[i,j,k]=min{dp[i,j,k-1],dp[i,k,k-1]+dp[k,j,k-1]}#include<iostream>#include<vector>#include<string>using namespace std;int main(){int dist[100][100], s = 0, n = 0, max = 0, start = 0, end = 0;string temp,result;vector<string> in;int path[100][100];memset(dist, 0, sizeof(int)*10000);memset(path, -1, sizeof(int)*10000);while (cin>>temp){in.push_back(temp);n++;}//初始化有向图for (int i = 0; i < n; i++){temp = in[i].substr(1,3);for (int j = 0; j < n; j++){if (i != j){if (in[j].substr(0, 3).compare(temp) == 0){dist[i][j] = 1;}}}}//动态规划实现弗洛伊德算法for (int k = 0; k < n; k++){for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (dist[i][k] != 0 && dist[k][j] != 0){if (dist[i][k] + dist[k][j]>dist[i][j]){dist[i][j] = dist[i][k] + dist[k][j];path[i][j] = k;}}}}}for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){if (max<dist[i][j]){max = dist[i][j];start = i;end = j;}}}temp = "";result = in[start][0];int mid = path[start][end];while (mid >= 0){temp = in[mid][0] + temp;mid = path[start][mid];}result += temp;result += in[end];cout << result << endl;return 0;}