Hdu 4786 Fibonacci Tree (minimum and maximum spanning Tree)

Question:

N vertices and M edges. Each edge is connected to two vertices u and v, and has a weight value of c, which is a non-zero value.

Ask if we can form N vertices into a spanning tree, and the edge weight of this tree is a number of ononacii. (Fibonacii number = 1, 2, 3, 5, 8 ....)

 

Ideas:

If a tree can be generated. The minimum spanning tree and maximum spanning tree are available. Assume that the minimum mst p and maximum mst q have been generated.

Replace P with one edge to get another spanning tree. The edge weight is 1 larger than the edge weight of P if it is not equal to P. (Because the edge value is non-zero, that is, one ). Similarly, we can get Q.

So all the values between the edge right of P and the edge right of Q can be obtained. Therefore, you can determine whether the number of fiber-ACII instances exists.

 

Code:

Struct node {int u, v, c;} edge [100005]; bool cmp (node a, node B) {return. c <B. c;} int fa [100005]; int T, n, m; int findFa (int x) {return fa [x] = x? X: fa [x] = findFa (fa [x]);} int kruskal1 () {rep (I, 1, n) fa [I] = I; int res = 0; rep (I, 1, m) {int fx = findFa (edge [I]. u); int fy = findFa (edge [I]. v); if (fx! = Fy) {fa [fx] = fy; res + = edge [I]. c ;}} int tx = findFa (1); rep (I, 2, n) if (findFa (I )! = Tx) return-1; return res;} int kruskal2 () {rep (I, 1, n) fa [I] = I; int res = 0; rep2 (I, m, 1) {int fx = findFa (edge [I]. u); int fy = findFa (edge [I]. v); if (fx! = Fy) {fa [fx] = fy; res + = edge [I]. c ;}} int tx = findFa (1); rep (I, 2, n) if (findFa (I )! = Tx) return-1; return res;} bool isFibo [100005]; void FiboD () {mem (isFibo, false); int a = 1, B = 2; isFibo [1] = isFibo [2] = true; for (;) {int t = a + B; a = B, B = t; if (t> 100000) break; isFibo [t] = true;} int main () {// freopen ("test. in "," r ", stdin); cin> T; FiboD (); rep (t, 1, T) {scanf (" % d ", & n, & m); rep (I, 1, m) scanf ("% d", & edge [I]. u, & edge [I]. v, & edge [I]. c); sort (edge + 1, edge + 1 + m, cmp); int mins = kruskal1 (); int maxs = kruskal2 (); printf ("Case # % d: ", t); if (mins =-1 | maxs =-1) puts (" No "); else {bool flag = false; rep (I, mins, maxs) if (isFibo [I]) {flag = true; break;} if (flag) puts ("Yes"); else puts ("No ");}} // fclose (stdin );}

 

Hdu 4786 Fibonacci Tree (minimum and maximum spanning Tree)