hdu 2614 BEAT 回溯的dfs
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Zty is a man that always full of enthusiasm. He wants to solve every kind of difficulty ACM problem in the world. And he has a habit that he does not like to solve a problem that is easy than problem he had solved. Now yifenfei give him n difficulty problems, and tell him their relative time to solve it after solving the other one. You should help zty to find a order of solving problems to solve more difficulty problem. You may sure zty first solve the problem 0 by costing 0 minute. Zty always choose cost more or equal time’s problem to solve.
30 0 01 0 11 0 030 2 21 0 11 1 050 1 2 3 10 0 2 3 10 0 0 3 10 0 0 0 20 0 0 0 0
324
Hint: sample one, as we know zty always solve problem 0 by costing 0 minute. So after solving problem 0, he can choose problem 1 and problem 2, because T01 >=0 and T02>=0. But if zty chooses to solve problem 1, he can not solve problem 2, because T12 < T01. So zty can choose solve the problem 2 second, than solve the problem 1.
//http://blog.csdn.net/libin56842/article/details/41909429#include<iostream> #include<cstring>using namespace std;const int maxn=1001;int ans=0;int n;int mp[maxn][maxn];int vis[maxn];void dfs(int pos,int len,int tm){ans=max(len,ans);if(len==n)return;for(int i=1;i<=n;i++){if(!vis[i]&&mp[pos][i]>=tm){vis[i]=1;dfs(i,len+1,mp[pos][i]);vis[i]=0;}}}int main(){while(cin>>n){ans=0;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)cin>>mp[i][j];vis[1]=1;dfs(1,1,0);cout<<ans<<endl;}return 0;}
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