HDU2833 WuKong Floyd

/*Floyd的应用用Floyd求出任意两点间的最短路径两个定理：1.所求的路径一定是一断连续的路径2.如果路径(x,y)是a->b的最短路径中的一段，则min(a,b) = min(a,x) + min(x,y) + min(y,b)最后只需找到同时在两条最短路径上，且距离最长的那一段*/#include <iostream>using namespace std;const int inf=200000000; int s[305][305];//记录最短路的长度int num[305][305];//同样长度的最短路的最长的跳数int main(){int n, m, i, j, k, a, b, c, d, ans, start, end, dis;while(scanf("%d%d", &n, &m), m || n){memset(num, 0, sizeof(num));for(i = 0; i <= n; i++){              for(j = 0; j <= n; j++)                 s[i][j]=(i==j)?0:inf;               }  for(i = 0; i < m; i++){scanf("%d%d%d", &start, &end, &dis);if(s[start][end] < dis)continue;s[start][end] = dis;s[end][start] = dis;num[start][end] = 1;num[end][start] = 1;}for(k=1;k<=n;k++){  for(i=1;i<=n;i++){  for(j=1;j<=n;j++){  if(s[i][j]>s[i][k]+s[k][j]){s[i][j]=s[i][k]+s[k][j];num[i][j] = num[i][k] + num[k][j];}else if (s[i][j] == s[i][k] + s[k][j] && num[i][j] < num[i][k] + num[k][j])num[i][j] = num[i][k] + num[k][j];}     }     }scanf("%d%d%d%d", &a, &b, &c, &d);ans = -1;for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){if(num[i][j] > ans && (s[a][b] == s[a][i] + s[i][j] + s[j][b]) && (s[c][d] == s[c][i] + s[i][j] + s[j][d]))ans = num[i][j];}}printf("%d\n", ans + 1);}return 0;}