Fzu 2090 travel Agency troubles Floyd to find the least ring of the graph without direction

Topic Link: The trouble of travel agency

Test instructions is the smallest loop of the graph, and if so, the number of outputs and the output weights.

Just mended a Floyd dynamic programming principle, using the idea of sliding arrays. So, this problem is the deformation of Floyd thought. In the process of k from 1 to n is updated to K, the Mindis array is saved only by 1~k-1 ordinal point, any two between the shortest path weights, this time, select Point K as the starting point of the ring as the end point, in [1, k) Select two points I, J to get a ring, the weight of the ring is Mindis[i ][J] + dis[i][k] + dis[j][k]. So the traversal is all the rings, and there are no duplicate points within the ring. Rings are not repeated. So ingenious ~ ~ to offer my knees ...

Attached code:

#include <stdio.h> #include <string.h> #include <iostream> #define INF 100000000using namespace std;    int Dis[110][110];int Mindis[110][110];int Main () {int t;    CIN >> T;        while (t--) {int n, m;        CIN >> n >> m;            for (int i=1, i<=n; ++i) {for (int j=1; j<=n; ++j) {dis[i][j] = inf;            }} for (int i=0; i<m; ++i) {int x, y, W;            Cin >> x >> y >> w;                if (Dis[x][y] > W) {dis[x][y] = W;            Dis[y][x] = W; }} for (int i=1, i<=n; ++i) {for (int j=1; j<=n; ++j) {mindis[i][j] = dis I            [j];        }} int ansdis = inf, anscnt = 0;                    for (int k=1, k<=n; ++k) {for (Int. I=1; i<k; ++i) {for (int j=i+1; j<k; ++j) {    if (Ansdis > Dis[i][k] + dis[j][k] + mindis[i][j]) {                    Ansdis = Dis[i][k] + dis[j][k] + mindis[i][j];                    anscnt = 1;                    } else if (Ansdis = = Dis[i][k] + dis[j][k] + mindis[i][j]) {anscnt++; }}} for (int i=1, i<=n; ++i) {for (int j=1; j<=n; ++J) {Mindis[i][j] = min (Mindis[i][j], mindis[i][k] + mindis[j][k]);//Feel the last mindis[j][k] if you switch to mindis[ K][J] The words should be WA's, however not.            So strange ...}        }} if (anscnt = = 0) {cout << " -1\n";    } else cout << ansdis << "<< anscnt << Endl; } return 0;}

Note: Mindis[j][k] should not be equal to mindis[k][j]. Puzzled...................

Fzu 2090 travel Agency troubles Floyd to find the least ring of the graph without direction