HDU 2489 Minimal Ratio Tree DFS enumeration point + minimum spanning tree belongs to the middle of the problem, double compare the size of time attention to accuracy

Minimal Ratio TreeTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2835 Accepted Submission (s): 841


Problem descriptionfor a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the foll Owing equation.




Given a complete graph of n nodes with all nodes and edges weighted, your task was to find a tree, which is a sub-graph of The original graph, with m nodes and whose ratio are the smallest among all the trees of M nodes in the graph.

Inputinput contains multiple test cases. The first line of all test case contains-integers n (2<=n<=15) and M (2<=m<=n), which stands for the numb Er of nodes in the graph and the number of nodes in the minimal ratio tree. Zeros end the input. The next line contains n numbers which stand for the weight of each node. The following n lines contain a diagonally symmetrical NXN connectivity matrix with each element shows the weight of the E Dge connecting one node with another. Of course, the diagonal would be all 0, since there are no edge connecting a node with itself.



All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) is integers and in the range of [1, 100].

The figure below illustrates the first test case in sample input. Node 1 and Node 3 form the minimal ratio tree.


Outputfor each test case output one line contains a sequence of the M nodes which constructs the minimal ratio tree. Nodes should is arranged in ascending order. If There is several such sequences, pick the one which has the smallest node number; If there ' s a tie, look at the second smallest node number, etc. Please note that the nodes is numbered from 1.
Sample Input

3 230 20 100 6 26 0 32 3 02 21 10 22 00 0

Sample Output

1 31 2

Reprint Please declare: http://blog.csdn.net/lionel_d
When writing code, always make SB's mistake!!! Cause wrong for a long time!! Always thought it was the wrong precision, O (╯-╰) O first enumerate m points, ask this m-point of the smallest spanning tree,, can,, at first I was the first to seek n points of the smallest spanning tree, and then enumerate m points from the tree,, the results are always wrong, and then look at other people's solution, only to understand their thinking wrong is very serious AH!!!
Look at the code:

#include <cstdio> #include <cstring> #include <algorithm> #define MAX 20#define INF 100000000using namespace std; int Graph[max][max], Lowcost[max], Node[max], Path[max], Temp[max];int N, m; bool CMP (const int A, con St int b) {return a<b;} int prim (int s) {bool Visited[max]; memset (visited,false,sizeof (visited)); int sum = 0; for (int i = 0; i < m; ++i) {Low Cost[temp[i]] = Graph[s][temp[i]];} Visited[s] = true; for (int i = 0; i < m-1; ++i) {int min = INF, index =-1; for (int j = 0; j < m; ++j) {if (!visit ED[TEMP[J]] && lowcost[temp[j]]<min) {min = lowcost[temp[j]]; index = Temp[j];}} if (index = =-1) {break;} Visited[index] = true; sum + = min; for (int j = 0; j < m; ++j) {if (!visited[temp[j]] && Lowcost[temp[j]]>gra PH[INDEX][TEMP[J]]) {Lowcost[temp[j]] = Graph[index][temp[j];}}} return sum;} Double ans = inf*1.0; void DFS (int num, int count) {if (count = = m) {int sumofedge = Prim (temp[0]), Sumofnode = 0;for (int i = 0; I < m; ++i){Sumofnode + = Node[temp[i]];} Double T = sumofedge*1.0/sumofnode, if (ans-t>0.00001) {ans = t; for (int i = 0; i < m; ++i) {path[i] = Temp[i];}} return;} for (int i = num+1; I <= n; ++i) {Temp[count] = i;D FS (i,count+1);}} int main () {while (~scanf ("%d%d", &n,&m) && (m| | N) {for (int i = 1; I <= n; ++i) {scanf ("%d", &node[i]);} for (int i = 1; I <= n; ++i) {for (int j = 1; j <= N; ++j) {scanf ("%d", &graph[i][j]);} Graph[i][i] = INF;} ans = inf*1.0; for (int i = 1; I <= n; ++i) {temp[0] = i;D FS (i,1);} Sort (path,path+m,cmp); for (int i = 0; i < m; ++i) {printf ("%d", path[i]), if (i! = m-1) {printf ("");}} Puts ("");} return 0;}

with June

HDU 2489 Minimal Ratio Tree DFS enumeration point + minimum spanning tree belongs to the middle of the problem, double compare the size of time attention to accuracy