• HDU 4277

    USACO ORZ

    Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 765 Accepted Submission(s): 253


    Problem Description
    Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.
    I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N fence segments and must arrange them into a triangular pasture. Ms. Hei must use all the rails to create three sides of non-zero length. Calculating the number of different kinds of pastures, she can build that enclosed with all fence segments.
    Two pastures look different if at least one side of both pastures has different lengths, and each pasture should not be degeneration.
     
    Input
    The first line is an integer T(T<=15) indicating the number of test cases.
    The first line of each test case contains an integer N. (1 <= N <= 15)
    The next line contains N integers li indicating the length of each fence segment. (1 <= li <= 10000)
     
    Output
    For each test case, output one integer indicating the number of different pastures.
     
    Sample Input
    1 3 2 3 4
     
    Sample Output
    1
     
     1 //爆搜+set判重 ,实际还是hash判重，map判重有两个参数 
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <set>
 5 using namespace std;
 6 
 7 
 8 int n;
 9 int arr[20];
10 set <long long> sset;
11 
12 void dfs(int a, int b, int c, int cur)
13 {
14     if (cur==n)
15     {
16         if (a>b||b>c||a>c) return;
17         if (a&&b&&c&&a+b>c)
18         {
19             long long t = 1000000000000LL*a+1000000LL*b+c;
20             sset.insert(t);
21         }
22         return;
23     }
24     dfs(a+arr[cur],b,c,cur+1);
25     dfs(a,b+arr[cur],c,cur+1);
26     dfs(a,b,c+arr[cur],cur+1);
27 }
28 
29 int main()
30 {
31      int i,j,k,T;
32      cin>>T;
33      while(T--)
34      {
35           sset.clear();
36           cin>>n;
37           for(i = 0 ; i < n ; i++ )
38                cin>>arr[i];
39           dfs(0,0,0,0);
40           cout<<sset.size()<<endl;
41      }
42 return 0;
43 }
     1 //wa,测试没过 
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <set>
 5 using namespace std;
 6 
 7 
 8 int n;
 9 int arr[20];
10 set <long long> sset;
11 
12 bool is_tir(int a, int b, int c)
13 {
14      int p = (a+b+c)>>1;
15      if(p>a&&p>b&&p>c)
16           return 1;
17      return 0;
18 }
19 
20 void dfs(int a, int b, int c, int cur)
21 {
22     if (cur==n)
23     {
24         if (is_tir(a,b,c))
25         {
26             long long t = 1000000000000LL*a+1000000LL*b+c;
27             sset.insert(t);
28         }
29         else
30           return;
31     }
32     dfs(a+arr[cur],b,c,cur+1);
33     dfs(a,b+arr[cur],c,cur+1);
34     dfs(a,b,c+arr[cur],cur+1);
35 }
36 
37 int main()
38 {
39      int i,j,k,T;
40      cin>>T;
41      while(T--)
42      {
43           sset.clear();
44           cin>>n;
45           for(i = 0 ; i < n ; i++ )
46                cin>>arr[i];
47           dfs(0,0,0,0);
48           cout<<sset.size()<<endl;
49      }
50 return 0;
51 }
     1  //ＡＣ ,自己写的 
 2  #include <iostream>
 3  #include <algorithm>
 4  #include <set>
 5  using namespace std;
 6  
 7  
 8  int n,sum;
 9  int arr[20];
10  set <long long> sset;
11  
12  bool is_tir(int a, int b, int c)
13  {    
14       int p = (sum+1)>>1;//必须加上1 ，想想2 3  4 
15       if(p>a&&p>b&&p>c)
16            return 1;
17       return 0;
18  }
19  
20  void dfs(int a, int b, int c, int cur)
21  {
22      if (cur==n)
23      {    
24          if (is_tir(a,b,c))
25          {
26              if(a>b||b>c||a>c)
27                return ;
28              long long t = 1000000000000LL*a+1000000LL*b+c;
29              sset.insert(t);
30          }
31          else
32            return;
33      }
34      else//ce（应该是下标越上界），加上else后就tle
35      {
36           dfs(a+arr[cur],b,c,cur+1);
37           dfs(a,b+arr[cur],c,cur+1);
38           dfs(a,b,c+arr[cur],cur+1);
39      }
40  }
41  
42  int main()
43  {
44       int i,j,k,T;
45       cin>>T;
46       while(T--)
47       {
48            sum = 0;
49            sset.clear();
50            cin>>n;
51            for(i = 0 ; i < n ; i++ )
52            {
53                cin>>arr[i];
54                sum += arr[i];
55            }
56            dfs(0,0,0,0);
57            cout<<sset.size()<<endl;
58       }
59  return 0;
60  }
    View Code 
     1  //TLE
 2  #include <iostream>
 3  #include <algorithm>
 4  #include <set>
 5  using namespace std;
 6  
 7  
 8  int n,sum;
 9  int arr[20];
10  set <long long> sset;
11  
12  bool is_tir(int a, int b, int c)
13  {    
14       int p = (sum+1)>>1;
15       if(p>a&&p>b&&p>c)
16            return 1;
17       return 0;
18  }
19  
20  void dfs(int a, int b, int c, int cur)
21  {
22      if (cur==n)
23      {    
24          if (is_tir(a,b,c))
25          {
26              if(a>b)
27                swap(a,b);
28              if(a>c)
29                swap(a,c);
30              if(b>c)
31                swap(b,c);
32              long long t = 1000000000000LL*a+1000000LL*b+c;
33              sset.insert(t);
34          }
35          else
36            return;
37      }
38      else//ce（应该是下标越上界），加上else后就tle
39      {
40           dfs(a+arr[cur],b,c,cur+1);
41           dfs(a,b+arr[cur],c,cur+1);
42           dfs(a,b,c+arr[cur],cur+1);
43      }
44  }
45  
46  int main()
47  {
48       int i,j,k,T;
49       cin>>T;
50       while(T--)
51       {
52            sum = 0;
53            sset.clear();
54            cin>>n;
55            for(i = 0 ; i < n ; i++ )
56            {
57                cin>>arr[i];
58                sum += arr[i];
59            }
60            dfs(0,0,0,0);
61            cout<<sset.size()<<endl;
62       }
63  return 0;
64  }
    //小熊的位运算枚举超时算法，还不太懂
// Note:Your choice is C++ IDE
#include <iostream>
#include <set>
#include <algorithm>
using namespace std;

struct Side{
    long s[16];
    long cnt;
    Side():cnt(0){memset(s, 0, sizeof s);}
};

long val[16];
long sid1, sid2, sid3;
long n;

set<long> st;
void Insert()
{
    long hashVal = sid1*100000000 + sid2*10000 + sid3;
    if(st.find(hashVal) == st.end())
    {
        st.insert(hashVal);
    }
}

int main()
{
//    freopen("sjx.in", "r", stdin);
    long i, j, t;
    long m, sum;
    cin>>t;
    while(t--)
    {
        st.clear();
        cin >> n;
        sum = 0;
        for(int ii=0; ii<n; ii++)
        {
            cin>>val[ii];
            sum += val[ii];
        }
        m = (1<<n) - 1;//n个1的二进制
        Side sd1;
        for(i=1; i<m; i++)
        {
            sd1.cnt = 0;
            sid1 = 0;
            for(long k=0; k<n; k++)
            {
                if((i>>k)&1)
                {
                    sd1.s[sd1.cnt++] = val[k];
                }
                else
                {
                    sid1 += val[k];
                }
            }
            double three = sum/3.0;
            if((double)sid1 >= three)
            {
                long n = (1<<sd1.cnt) - 1;
                for(j=1; j<n; j++)
                {
                    sid2 = sid3 = 0;
                    for(long k=0; k<sd1.cnt; k++)
                    {
                        if((j>>k)&1)
                        {
                            sid2 += sd1.s[k];
                            if(sid2 > sid1) goto Imp;
                        }
                        else
                        {
                            sid3 += sd1.s[k];
                            if((double)sid3 > three) goto Imp;
                        }
                    }
                    if(sid3 <= sid2 
                        //&& sid2 <= sid1
                        && sid3+sid2 > sid1)
                    Insert();
                    Imp:;
                }
            }
        }
        cout<<st.size()<<endl;//<<","<<clock()<<endl;
    }
    //while(1);
    return 0;
}
  • 相关阅读:
    jquery源码分析
    面试笔试整理3：深度学习机器学习面试问题准备（必会）
    深度学习岗位面试问题整理笔记
    常见面试之机器学习算法思想简单梳理
    TensorFlow与OpenCV，读取图片，进行简单操作并显示
    payload
    IPMB接口协议总结
    uboot 网络驱动模型
    对于phy芯片的认识
    网口扫盲三:以太网芯片MAC和PHY的关系
  • 原文地址：https://www.cnblogs.com/hxsyl/p/2677281.html
Copyright © 2020-2023  润新知