HDU 2614 Beat(DFS)
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.
Each test case include, first one integer n ( 2< n < 15).express the number of problem.
Than n lines, each line include n integer Tij ( 0<=Tij<10), the i’s row and j’s col integer Tij express after solving the problem i, will cost Tij minute to solve the problem j.
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.
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
//#define LOCAL
int pro[][],maxx,n;
bool vis[];
void dfs(int p,int las,int cnt)
{
int flag=;
for(int i=; i<=n; i++)
{
if(!vis[i]&&(pro[p][i]>=las))
{
vis[i]=;
dfs(i,pro[p][i],cnt+);
vis[i]=;
flag=;
}
}
if(!flag)if(cnt>maxx)maxx=cnt;
}
int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
//Start
while(cin>>n)
{
maxx=;
memset(pro,,sizeof pro);
memset(vis,,sizeof vis);
for(int i=; i<=n; i++)
for(int j=; j<=n; j++)
cin>>pro[i][j];
vis[]=;
dfs(,,);
printf("%d\n",maxx);
}
return ;
}
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