hdu 2489 Minimal Ratio Tree (DFS枚举+MST)

参考链接:http://blog.csdn.net/xingyeyongheng/article/details/9373271

     http://www.cnblogs.com/chenxiwenruo/p/3294668.html

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<map>
 9 #include<iomanip>
10 #define INF 99999999
11 using namespace std;
12 
13 const int MAX = 15 + 5;
14 int edge[MAX][MAX], dist[MAX], node[MAX];//node记录最终选的点 
15 int vale[MAX], temp[MAX], n, m;//temp记录选的m个点
16 double minratio;
17 bool mark[MAX];
18 
19 int Prim(int s){
20     int sum = 0;
21     for (int i = 1; i <= m; ++i)mark[temp[i]] = false, dist[temp[i]] = edge[s][temp[i]];
22     mark[s] = true;
23     dist[s] = 0;
24     for (int i = 1; i<m; ++i){
25         int point = s;
26         for (int j = 1; j <= m; ++j){
27             if (point == s && !mark[temp[j]])point = temp[j];
28             if (!mark[temp[j]] && dist[point]>dist[temp[j]])point = temp[j];
29         }
30         mark[point] = true;
31         sum += dist[point];
32         for (int j = 1; j <= m; ++j){
33             if (!mark[temp[j]] && edge[point][temp[j]]<dist[temp[j]])dist[temp[j]] = edge[point][temp[j]];
34         }
35     }
36     return sum;
37 }
38 /*
39 dfs枚举m个点,num代表目前选取了多少个点,k代表第num个点为k。
40 表示前num个点选自1~k,剩余的点从k+1~n中选。
41 */
42 void dfs(int u, int num){
43     if (num == m){
44         int ans = 0;
45         for (int i = 1; i <= m; ++i)ans += vale[temp[i]];
46         double sum = Prim(u)*1.0 / ans;
47         if (sum - minratio < -(1e-9)){
48             minratio = sum;
49             for (int i = 1; i <= m; ++i) node[i] = temp[i];
50         }
51         return;
52     }
53  //若剩余的点的个数(n-k)加上目前选取的个数num小于m的话,说明即使接下来n-k个点都选取,也选不足m个点,直接return
54     if (n - u + num<m)return;
55     for (int i = u + 1; i <= n; ++i){
56         temp[num + 1] = i;
57         dfs(i, num + 1);
58     }
59 }
60 
61 int main(){
62     while (cin >> n >> m, n + m){
63         minratio = INF*1.0;
64         for (int i = 1; i <= n; ++i)cin >> vale[i];
65         for (int i = 1; i <= n; ++i){
66             for (int j = 1; j <= n; ++j){
67                 cin >> edge[i][j];
68             }
69         }
70         for (int i = 1; i <= n; ++i){
71             temp[1] = i;
72             dfs(i, 1);
73         }
74         for (int i = 1; i<m; ++i)cout << node[i] << ' ';
75         cout << node[m] << endl;
76     }
77     return 0;
78 }

别的大牛DFS写法

 1 //调用时:dfs(1,0,0);
 2 
 3 //dep表示点的编号,cnt表示选取的点的个数,sum_pw表示目前选取了cnt个点后总的点权值
 4 void dfs(int dep, int cnt, int sum_pw) {
 5     if(cnt == m) {
 6         ...;
 7         return ;
 8     }
 9     if(dep == n + 1) return ;
10     use[dep] = true;  //选取点dep,这里use[i]=true表示选取点i,在用prim求最小生成树的时候有用
11     dfs(dep + 1, cnt + 1, sum_pw + weight[dep]);
12     use[dep] = false; //不选取点dep
13     dfs(dep + 1, cnt, sum_pw);
14 }
原文地址:https://www.cnblogs.com/usedrosee/p/4249601.html