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Uva LA6450 Social Advertising DFS
You have decided to start up a new social networking company. Other existing popular social networks
already have billions of users, so the only way to compete with them is to include novel features no
other networks have.
Your company has decided to market to advertisers a cheaper way to charge for advertisements (ads).
The advertiser chooses which users‘ \wall" the ads would appear on, and only those ads are charged.
When an ad is posted on a user‘s wall, all of his/her friends (and of course the user himself/herself )
will see the ad. In this way, an advertiser only has to pay for a small number of ads to reach many
more users.
You would like to post ads to a particular group of users with the minimum cost. You already have
the \friends list" of each of these users, and you want to determine the smallest number of ads you have
to post in order to reach every user in this group. In this social network, if A is a friend of B, then B
is also a friend of A for any two users A and B.
Input
The input consists of multiple test cases. The rst line of input is a single integer, not more than
10, indicating the number of test cases to follow. Each case starts with a line containing an integer n
(1 n 20) indicating the number of users in the group. For the next n lines, the ith line contains the
friend list of user i (users are labelled 1; : : : ; n). Each line starts with an integer d (0 d < n) followed
by d labels of the friends. No user is a friend of himself/herself.
Output
For each case, display on a line the minimum number of ads needed to be placed in order for them to
reach the entire group of users.
Sample Input
2
5
4 2 3 4 5
4 1 3 4 5
4 1 2 4 5
4 1 2 3 5
4 1 2 3 4
5
2 4 5
2 3 5
1 2
2 1 5
3 1 2 4
Sample Output
1
2
题意：有个公司做广告，如果在某个人那里做广告，和他相连的朋友都可以知道广告，要求所有的人知道广告，并且做广告的数量最少。
所以是找最少的点把所有相连的点覆盖。
思路：枚举任意一种组合当只有一个人时是否可以全部覆盖，如果不行，就枚举任意两个人是否可以全部覆盖，再依次类推，知道找到就break；
DFS到达叶子节点时才计算是否全部覆盖，这个题很容易超时。后来看看别人是怎么写的，他用了2进制枚举，那样时间复杂度就更高了，
不过他没有用邻接表，而是用二进制建边，有点6。
本人代码：
#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn=25;int n,flag;vector<int> f[maxn];int vis[maxn];int cnt[maxn];bool is_ok(int s){ memset(vis,0,sizeof(vis)); int i,j,sum=0; for(i=0;i<s;i++) { if(!vis[cnt[i]]) { vis[cnt[i]]=1;sum++; } for(j=0;j<f[cnt[i]].size();j++) { if(!vis[f[cnt[i]][j]]) { vis[f[cnt[i]][j]]=1;sum++; } } } if(sum==n) return 1; return 0;}void dfs(int now,int s,int dep){ if(now>n+1) return ; if(s==dep) { if(is_ok(s)) flag=1; return ; } cnt[s]=now; dfs(now+1,s+1,dep); dfs(now+1,s,dep);}int main(){ int t,i,k,p; scanf("%d",&t); while(t--) { scanf("%d",&n); for(i=1;i<=n;i++) f[i].clear(); for(i=1;i<=n;i++) { scanf("%d",&k); while(k--) { scanf("%d",&p); f[i].push_back(p);f[p].push_back(i); } } flag=0; for(i=1;i<=n;i++) { dfs(1,0,i); if(flag) break; } printf("%d\n",i); } return 0;}
用他的建边代码改进，时间少了100+ms
改进代码：
#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn=25;int n,flag;int vis[maxn];int cnt[maxn];int eg[maxn];bool is_ok(int s){ int ans = 0; for(int i=0;i<s;i++) ans |=eg[cnt[i]]; if(ans == (1<<n)-1 ) return true; //所有顶点都访问了 return false;}void dfs(int now,int s,int dep){ if(now>n+1) return ; if(s==dep) { if(is_ok(s)) flag=1; return ; } cnt[s]=now; dfs(now+1,s+1,dep); dfs(now+1,s,dep);}int main(){ int t,i,k,p; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(eg,0,sizeof(eg)); for(i=1;i<=n;i++) { scanf("%d",&k); eg[i] |=1<<(i - 1); while(k--) { scanf("%d",&p); eg[i] |=1<<(p - 1); //表示第i条边与第p条边有连边 } } flag=0; for(i=1;i<=n;i++) { dfs(1,0,i); if(flag) break; } printf("%d\n",i); } return 0;}/*254 2 3 4 54 1 3 4 54 1 2 4 54 1 2 3 54 1 2 3 452 4 52 3 51 22 1 53 1 2 4 */