zoukankan      html  css  js  c++  java
  • hdu2614

                                         

     

    Beat

    Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 374    Accepted Submission(s): 242


    Problem Description
    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.
     
    Input
    The input contains multiple test cases.
    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.
     
    Output
    For each test case output the maximum number of problem zty can solved.


     
    Sample Input
    3
    0 0 0
    1 0 1
    1 0 0
    3
    0 2 2
    1 0 1
    1 1 0
    5
    0 1 2 3 1
    0 0 2 3 1
    0 0 0 3 1
    0 0 0 0 2
    0 0 0 0 0
     
    Sample Output
    3 2 4
    Hint
    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<iostream>
    #include<cstring>
    #include<string>
    using namespace std;
    int map[21][21],visit[21];
    int max1,n;
    void dfs(int x,int y,int ans)
    {
        int i;
        int flag = 0;
        for(i=1;i<=n;i++)
        {
            if(!visit[i] && map[x][i] >= y)
            {
                visit[i] = 1;
                dfs(i,map[x][i],ans+1);
                visit[i] = 0;
                flag = 1;
            }
        }
        if(!flag)
        {
            if(ans>max1)
            {
                max1 = ans;
                
            }
        }
        return ;
    }
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            int i , j ;max1 = 0;
            memset(visit,0,sizeof(visit));
            memset(map,0,sizeof(map));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    scanf("%d",&map[i][j]);
                }
            }
    
            visit[1] =1;
            dfs(1,0,1);
            printf("%d
    ",max1);
        }
        return 0;
    }
     
     
     
     
     
     
  • 相关阅读:
    Minimum Path Sum
    Unique Paths II
    Unique Paths
    Rotate List
    Permutation Sequence
    Merge Intervals
    Jump Game
    Group Anagrams
    Combination Sum II
    评分
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/3232726.html
Copyright © 2011-2022 走看看