UVa 10985 Rings ' n ' Ropes floyd+bfs

n rings and a rope with a length of 1 m

Ask the right hand to hold any 2 rings up to a few ropes straightened

Hold which 2 don't know so 22 enum

For 2 rings You can consider the shortest possible short circuit and then there may be a number of the shortest lines in the path statistics.

I know how to get out of a sample.

Example 3 is a ring pull 0 and 2 you have 2 of the shortest, each one is 3.

Example 4 is the same

#include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std;
const int MAX = 130;
int A[max][max];
int Map[max][max];
BOOL Vis[max][max];
int n,m;
int cnt;
	void Floyd () {int i,j,k; for (k = 0, k < n; k++) for (i = 0, i < n; i++) for (j = 0; J < N; j + +) A[i][j] = min (a[i][j],a[i][k]+a[k][
J]);
	} void BFs (int s,int e) {queue <int> q;
	Q.push (s);
	int i;
		while (!q.empty ()) {int u = q.front ();
		Q.pop ();
			for (i = 0; i < n; i++) {if (vis[u][i]) continue;
			if (!map[u][i]) continue;
				if (A[e][i] + 1 = = A[e][u]) {cnt++;
				Vis[u][i] = Vis[i][u] = true;
			Q.push (i);
	}}}} int main () {int t,cas = 1;
	int i,j;
	int s,e;
	scanf ("%d", &t);
		while (t--) {memset (a,0,sizeof (a));
		memset (map,0,sizeof (map));
		scanf ("%d%d", &n,&m);
				for (i = 0; i < n; i++) for (j = 0; J < N; j + +) {if (i = = j) A[i][j] = 0;
		else a[i][j] = 999999999;	} for (i = 0; i < m; i++) {scanf ("%d%d", &s,&e);
			A[s][e] = A[e][s] = 1;
		Map[s][e] = Map[e][s] = 1;
		} Floyd ();
		int max = 0;
				
				for (i = 0, i < n; i++) {for (j = i+1; J < N; j + +) {if (a[i][j] = = 999999999) continue;
				memset (vis,false,sizeof (VIS));
				CNT = 0;
				BFS (I,J);
			if (Max < CNT) max = CNT;
		}}//printf ("%d%d\n", a[0][1],a[1][0]);
	printf ("Case #%d:%d\n", Cas++,max);
} return 0; }