Minimal Ratio Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
【Problem Description】
For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation.
【Input】
Input contains multiple test cases. The first line of each test case contains two integers n (2<=n<=15) and m (2<=m<=n), which stands for the number of nodes in the graph and the number of nodes in the minimal ratio tree. Two 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 n×n connectivity matrix with each element shows the weight of the edge connecting one node with another. Of course, the diagonal will be all 0, since there is 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) are 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.
All the weights of both nodes and edges (except for the ones on the diagonal of the matrix) are 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.
【Output】
For each test case output one line contains a sequence of the m nodes which constructs the minimal ratio tree. Nodes should be arranged in ascending order. If there are 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 are numbered from 1 .
【Sample Input】
3 2 30 20 10 0 6 2 6 0 3 2 3 0 2 2 1 1 0 2 2 0 0 0
【Sample Output】
1 3 1 2
【题意】
给出一张n个点的图，图中的每一个结点以及每一条边都有其权值，要求从中选出m个点，找到m-1条边将其连接，使得边权值与点权值的比值达到最小。
【分析】
要使得比值最小，则点权值和尽可能地大同时边权值和尽可能地小。直接上考虑，边权值和尽可能小即对这m个点作最小生成树。
而题目给定的n不大，故可以用DFS搜出需要的m个点，然后对m个点进行最小生成树，中间注意判断和保存即可。
我用了一个dijkstra+优先队列的prim去找MST，这也是我第一次尝试使用STL的优先队列。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 6 using namespace std; 7 8 bool flag[16],outp[16]; 9 int n,m,node[16]; 10 int ma[16][16]; 11 double mi; 12 13 typedef struct heaptyp 14 { 15 int num,key; 16 friend bool operator < (heaptyp a,heaptyp b) 17 { 18 return a.num>b.num; 19 } 20 } heaptype; 21 22 void prim(int s,int tot) 23 { 24 int i,j,now,ans; 25 bool fla[16]; 26 priority_queue<heaptype>heap; 27 heaptype aaa; 28 29 memset(fla,0,sizeof(fla)); 30 fla[s]=true; 31 ans=0; 32 for (i=1;i<=n;i++) 33 if (flag[i]&&ma[s][i]) 34 { 35 heaptype temp; 36 temp.num=ma[s][i]; 37 temp.key=i; 38 heap.push(temp); 39 aaa=heap.top(); 40 } 41 42 for (j=1;j<m;j++) 43 { 44 heaptype h=heap.top(); 45 heap.pop(); 46 aaa=heap.top(); 47 while (fla[h.key]) 48 { 49 h=heap.top(); 50 heap.pop(); 51 } 52 now=h.key; 53 fla[now]=true; 54 ans+=h.num; 55 for (i=1;i<=n;i++) 56 if (flag[i]&&ma[now][i]) 57 if (!fla[i]) 58 { 59 heaptype temp; 60 temp.num=ma[now][i]; 61 temp.key=i; 62 heap.push(temp); 63 aaa=heap.top(); 64 } 65 } 66 double rat=(double)ans/tot; 67 if (mi-rat>0.0000001) 68 { 69 mi=rat; 70 for (int i=1;i<=n;i++) outp[i]=fla[i]; 71 } 72 } 73 74 void dfs(int now,int last,int tot) 75 { 76 if (now==m) 77 { 78 int i; 79 for (i=1;i<=n;i++) 80 if (flag[i]) break; 81 prim(i,tot); 82 } 83 else 84 { 85 now++; 86 for (int i=last+1;i<=n-m+now;i++) 87 { 88 flag[i]=true; 89 dfs(now,i,tot+node[i]); 90 flag[i]=false; 91 } 92 } 93 } 94 95 int main() 96 { 97 scanf("%d%d",&n,&m); 98 while (!(n==0&&m==0)) 99 { 100 for (int i=1;i<=n;i++) scanf("%d",&node[i]); 101 for (int i=1;i<=n;i++) 102 for (int j=1;j<=n;j++) scanf("%d",&ma[i][j]); 103 104 mi=2147483647; 105 memset(flag,0,sizeof(flag)); 106 for (int i=1;i<=n-m+1;i++) 107 { 108 flag[i]=true; 109 dfs(1,i,node[i]); 110 flag[i]=false; 111 } 112 113 int i; 114 for (i=1;i<=n;i++) 115 if (outp[i]) 116 { 117 printf("%d",i); 118 break; 119 } 120 for (int j=i+1;j<=n;j++) 121 if (outp[j]) printf(" %d",j); 122 printf(" "); 123 scanf("%d%d",&n,&m); 124 } 125 126 return 0; 127 }