C++博客-我要一步一步往上爬!~~加油~加油~加油!-随笔分类-动态规划

pku + 1887

haozi — Thu, 10 Jun 2010 04:14:00 GMT

水题一道！

1 #include <iostream>
2 #include <vector>
3 #include <new>
4 using namespace std;
5
6 struct nod {
7     int num;
8     int max_num;
9 };
10
11 int m = 1;
12
13 void ji_suan( vector<nod *> &v2 )
14 {
15     int max = 1;
16     for ( vector<nod *> ::iterator p = v2.begin(),q = v2.end(); p != q; ++p )
17     {
18         for ( vector <nod *> ::iterator p1 = p + 1; p1 != q; ++p1 )
19         {
20             if ( (*p)->num > (*p1)->num )
21             {
22                 if ( (*p)->max_num + 1 > (*p1)->max_num )
23                 {
24                     (*p1)->max_num = (*p)->max_num + 1;
25                     if ( max < (*p1)->max_num )
26                         max = (*p1)->max_num;
27                 }
28             }
29         }
30     }
31     printf ( "Test #%d:\n",m++ );
32     cout << "  maximum possible interceptions: " << max << endl;
33 }
34
35 int main()
36 {
37     //freopen( "in.txt","r",stdin );
38     vector<nod *> v1;
39     int num;
40     while ( cin >> num && num != -1 )
41     {
42         int n;
43         nod *p = new nod;
44         p->max_num = 1;
45         p->num = num;
46         v1.push_back(p);
47         while ( cin >> n && n != -1 )
48         {
49             p = new nod;
50             p->num = n;
51             p->max_num = 1;
52             v1.push_back(p);
53         }
54         ji_suan( v1 );
55         v1.clear();
56         cout << endl;
57     }
58     return 0;
59 }

haozi 2010-06-10 12:14 发表评论

pku + 1989 最短非子序列

haozi — Wed, 09 Jun 2010 13:39:00 GMT

假设有排序的个数为n,则其最短的非子序长度，可由下面方法求得。

如果存在其有包括全排序的一个最小组合，(1,2,...n),(次序不限，只要有1...n的数字就行了,也有可能有重复数字)

找到其所有这样组合。可合成()()()...()[],( 设有k个括号,[]里的元素可能为空)
则其最短的非子序长度为k+1;
原因：
长度为k的子序,都可由前k个括号里抽出一个数组成。
但对于k+1,就找不到一个组合,因为最后一个组合里少了需要的最后一个数。

如本例子中：
14 5
1
5
3
2
5
1
3
4
4
2
5
1
2
3

首先有(1,5,3,2,5,1,3,4)(4,2,5,1,2,3)[]
故其最短非子序长度为3;

1 #include <iostream>
2 #include <new>
3 #include <vector>
4 #include <algorithm>
5 using namespace std;
6
7 int main()
8 {
9     //freopen( "in.txt","r",stdin );
10     int n,k;
11     cin >> n;
12     cin >> k;
13     vector<bool> v1( k+1,false );
14     int num = 1;
15     for ( int i = 0,j = 0,k1; i != n; ++i )
16     {
17         cin >> k1;
18         if ( v1[k1] == false )
19         {
20             ++j;
21             v1[k1] = true;
22         }
23         if ( j == k )
24         {
25             j = 0;
26             ++num;
27             for ( int k2 = 0; k2 != v1.size(); ++k2 )
28                 v1[k2] = false;
29         }
30     }
31     cout << num << endl;
32
33 }
34
35
36
37
38

haozi 2010-06-09 21:39 发表评论

pku + 2029

haozi — Tue, 08 Jun 2010 08:07:00 GMT

1 #include <iostream>
2 #include <string>
3 #include <new>
4 #include <vector>
5 #include <algorithm>
6 using namespace std;
7
8 int main()
9 {
10     //freopen( "in.txt","r",stdin );
11     int tree[100][100];
12     int sum[100][100];
13     int num;
14     while ( cin >> num && num != 0 )
15     {
16         int w,h;
17         cin >> w >> h;
18         for ( int i = 0; i != h; ++i )
19             for ( int j = 0; j != w; ++j )
20             {
21                 tree[i][j] = 0;
22                 sum[i][j] = 0;
23             }
24         int x,y;
25         for ( int i = 0; i != num; ++i )
26         {
27             cin >> x >> y;
28             tree[y-1][x-1] = 1;
29         }
30         //for ( int i = 0; i != h; ++i )
31         //{
32             //for ( int j = 0; j != w; ++j )
33                 //cout << tree[i][j] << " ";
34             //cout << endl;
35         //}
36         cin >> x >> y; //x是宽，y 是长
37         for ( int i = 0; i != h; ++i ) //对于宽度的叠加
38             for ( int j = 0; j != w - x + 1; ++j )
39                 for ( int k = j,n = 0; n != x; ++k,++n )
40                     sum[i][j] += tree[i][k];
41         int max1 = 0;
42         int s = 0;
43         for ( int i = 0; i != w - x + 1; ++i ) // 宽 // 对于长度的叠加
44             for ( int j = 0; j != h - y + 1; ++j ) // 高
45             {
46                 s = 0;
47                 for ( int k = j,n = 0; n != y; ++k,++n )
48                     s +=sum[k][i];
49                 max1 = max( s,max1 );
50             }
51         //for ( int i = 0; i != h; ++i )
52         //{
53             //for ( int j = 0; j != w-x + 1; ++j )
54                 //cout << sum[i][j] << " ";
55             //cout << endl;
56         //}
57         cout << max1 << endl;
58     }
59     return 0;
60 }
61
62
63
64

haozi 2010-06-08 16:07 发表评论

pku + 2033

haozi — Mon, 07 Jun 2010 13:01:00 GMT

1 #include <iostream>
2 #include <string>
3 #include <new>
4 #include <vector>
5 using namespace std;
6
7 inline int ji_suan( char &ch1,char &ch2 )
8 {
9     int j = ( ch1 - '0' ) * 10 + ( ch2 - '0' );
10     return j;
11 }
12 int main()
13 {
14     string str;
15     vector<int> v1;
16     while ( cin >> str && str != "0" )
17     {
18         for ( int i = 0; i != str.size(); ++i )
19             v1.push_back(0);
20         if ( str[0] - '0' >= 1 && str[0] - '0' <= 26 )
21             v1[0] = 1;
22         int i = 1;
23         for ( ; i != str.size(); ++i )
24         {
25             if ( ( i + 1 != str.size() && str[i+1] == '0' ) || str[i] == '0' || str[i-1] == '0' ) //判断 0 的情况
26             {
27                 v1[i] = v1[i-1];
28                 continue;
29             }
30             int j = ji_suan( str[i-1],str[i] );
31             if ( i == 1 && ( j >= 1 && j <= 26 ) ) // 然后是一个推的过程动动笔就明白了
32                 v1[i] = v1[i-1] + 1;
33             else if ( i > 1 )
34             {
35                 if ( j >= 1 && j <= 26 )
36                     v1[i] = v1[i-1] + v1[i-2];
37                 else v1[i] = v1[i-1];
38             }
39             else v1[i] = v1[i-1]; // 主要是针对 i == 1 不成立的时候
40         }
41         cout << v1[i-1] << endl;
42         v1.clear();
43     }
44     return 0;
45 }
46

haozi 2010-06-07 21:01 发表评论

pku + 2181

haozi — Sun, 06 Jun 2010 13:56:00 GMT

     郁闷!! 这道题其实是一道很水的! 当然是我明白以后才认识到的!
       这道题, 其实所有的要不知道道奇偶时间是吃还是不吃的好,所以开一个三维数组分别表示分别表示奇数偶数吃和不吃!  主要是判断上一次是吃好还是不吃的好!
       还有今天学会三维数组用法!! 咳我还有很长一段路要走! 加油!!!

1 1 #include <iostream>
2  2 #include <vector>
3  3 #include <string>
4  4 #include <new>
5  5 #include <algorithm>
6  6 using namespace std;
7  7
8  8 int s1[150001][2][2];
9  9
10 10
11 11 int main()
12 12 {
13 13     int s2[150001];
14 14     int num;
15 15     cin >> num;
16 16     for ( int i = 0; i != num; ++i )
17 17         scanf ( "%d",&s2[i] );
18 18     for ( int i = 1; i <= num; ++i )
19 19     {
20 20         s1[i][0][0] = max( s1[i-1][0][0],s1[i-1][0][1] ); // 偶数时间没有吃
21 21         s1[i][0][1] = max( s1[i-1][1][0],s1[i-1][1][1] ) - s2[i-1]; // 偶数时间吃了
22 22         s1[i][1][0] = max( s1[i-1][1][0],s1[i-1][1][1] ); // 奇数时间不吃
23 23         s1[i][1][1] = max( s1[i-1][0][0],s1[i-1][0][1] ) + s2[i-1];
24 24     }
25 25     cout << max( max( s1[num][0][0],s1[num][0][1] ),max( s1[num][1][0],s1[num][1][1] )  ) << endl;
26 26 }
27 27

haozi 2010-06-06 21:56 发表评论

pku + 2424

haozi — Thu, 03 Jun 2010 15:06:00 GMT

1 #include <iostream>
2 #include <vector>
3 #include <new>
4 #include <queue>
5 using namespace std;
6
7 bool pan_duan( int &n,queue<int> &v4,int &times )
8 {
9     if ( n != 0 )
10     {
11         v4.push(times + 30); // 直接进取吃了 30 分钟
12         --n;
13         return true;
14     }
15     else
16     {
17         int time2 = v4.front();
18         if ( times >= time2 )
19         {
20             v4.pop();
21             v4.push( times + 30 );
22             return true;
23         }
24         else if ( times + 30 >= time2  )
25         {
26             v4.pop();
27             v4.push( time2 + 30 );
28             return true;
29         }
30         else return false;
31     }
32 }
33
34
35 int main()
36 {
37     //freopen("in.txt","r",stdin  );
38     int a,b,c;
39     while ( scanf ( "%d",&a ) )
40     {
41         scanf ( "%d",&b );
42         scanf ( "%d",&c );
43         if ( a == 0 && b == 0 && c == 0 )
44             break;
45         queue<int> v1; //1~2
46         queue<int> v2; //3~4;
47         queue<int> v3; //5~6;
48         int t,t1,s,n;
49         char ch,ch1;
50         int num = 0;
51         bool y;
52         while ( 1 )
53         {
54             cin >> ch1;
55             if ( ch1 == '#' )
56                 break;
57             else cin >> t1 >> ch >> s >> n;
58             t1 = ( ch1 - '0' ) * 10 + t1;
59             int time;
60             time = t1 * 60 + s; // 等待
61             //cout << time << "fen" << endl;
62             if ( n >= 1 && n <= 2 )
63                 y = pan_duan( a,v1,time );
64             else if ( n >= 3 && n <= 4 )
65                 y = pan_duan( b,v2,time );
66             else if ( n >= 5 && n <= 6 )
67                 y = pan_duan( c,v3,time );
68             if ( y == true )
69                 num += n;
70         }
71         cout << num << endl;
72     }
73     return 0;
74 }
75
76

haozi 2010-06-03 23:06 发表评论

pku + 2533

haozi — Thu, 08 Apr 2010 14:46:00 GMT

这道题思路简单!

1 #include <cstdio>
2 #include <algorithm>
3 using namespace std;
4
5 int main()
6 {
7     int s1[1000];
8     int s2[1000];
9     int m;
10     scanf ( "%d",&m );
11     for ( int i = 0; i != m; ++i )
12     {
13         scanf ( "%d",&s1[i] );
14         s2[i] = 1;
15     }
16     int max1 = 0;
17     int max2;
18     for ( int i = 0; i != m; ++i )
19     {
20         max2 = 0;
21         for ( int j = 0; j != i; ++j )
22         {
23             if ( s1[i] > s1[j] )
24                 max2 = max( max2,s2[j] );
25         }
26         s2[i] += max2;
27         if ( s2[i] > max1 )
28             max1 = s2[i];
29     }
30     printf ( "%d\n",max1 );
31 }

haozi 2010-04-08 22:46 发表评论

pku + 1458 求最大公共子序列

haozi — Sun, 04 Apr 2010 06:44:00 GMT

1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 int main()
7 {
8     char s1[501];
9     char s2[501];
10     while ( scanf ( "%s",s1 ) != EOF ){
11     scanf ( "%s",s2 );
12     int s3[500][500];
13     int l1,l2;
14     l1 = strlen( s1 );
15     l2 = strlen( s2 );
16     for ( int i = 0; i != l1+1; ++i )
17         for ( int j = 0; j != l2 + 1; ++j  )
18             s3[i][j] = 0;
19     int max1 = 0;
20     for ( int i = 0; i != l1; ++i )
21         for ( int j = 0; j != l2; ++j )
22         {
23             if ( s1[i] == s2[j] )
24             {
25                 s3[i+1][j+1] = s3[i][j] + 1;
26                 if ( s3[i+1][j+1] > max1 )
27                     max1 = s3[i+1][j+1];
28             }
29             else
30                 s3[i+1][j+1] = max (s3[i+1][j], s3[i][j+1]);
31         }
32     printf ( "%d\n",max1 );
33     }
34     return 0;
35 }
36

这题其实不难,思想很简单,就是记录下前一个字母的最长子序列!
哈哈~~虽然参考书后还是相比较久,但总算会了!

haozi 2010-04-04 14:44 发表评论

pku + 1953

haozi — Thu, 01 Apr 2010 14:32:00 GMT

这题简单到不行,这题教育我,不要被方法圈住,有时候多动动笔,往后面写,规律就出来了!

1 #include <cstdio>
2
3 int main()
4 {
5     int n;
6     int c = 0;
7     scanf ( "%d",&n );
8     long s1[47];
9     while ( n != 0 )
10     {
11         int m;
12         scanf ( "%d",&m );
13         s1[0] = 1;
14         s1[1] = 2;
15         for ( int i = 2; i < m+1; ++i )
16             s1[i] = s1[i-1] + s1[i-2];
17         printf ( "Scenario #%d:\n",++c );
18         printf ( "%d\n",s1[m] );
19         printf ( "\n" );
20         --n;
21     }
22     return 0;
23 }
24
25
26

haozi 2010-04-01 22:32 发表评论

pku + 1163

haozi — Tue, 23 Mar 2010 14:17:00 GMT

1 #include <cstdio>
2 #include <algorithm>
3 using namespace std;
4
5 int main()
6 {
7     //freopen( "in.txt","r",stdin );
8     int s1[100][100];
9     int n;
10     scanf ( "%d",&n );
11     for ( int i = 0; i != n; ++i )
12         for ( int j = 0; j <= i; ++j )
13             scanf ( "%d",&s1[i][j] );
14     int max1[100];
15     for ( int i = 0; i != n; ++i )
16         max1[i] = s1[n-1][i];
17     //for ( int i = 0; i != n; ++i )
18     //    printf ( "%d ",max[i] );
19     for ( int i = n - 2;i >= 0; --i  )
20     {
21         int sum1 = 0;
22         int sum2 = 0;
23         int k = 0;
24         for ( int j = 0; j != i+1; ++j )
25         {
26             sum1 = s1[i][j] + max1[k];
27             sum2 = s1[i][j] + max1[k+1];
28             max1[k] = max( sum1,sum2 );
29             ++k;
30         }
31     }
32     printf ( "%d\n",max1[0] );
33 }

haozi 2010-03-23 22:17 发表评论

pku + 1157

haozi — Tue, 23 Mar 2010 08:28:00 GMT

1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 int main()
7 {
8     //freopen( "in.txt","r",stdin );
9     int s1[100][100];
10     int f,v;
11     scanf ( "%d%d",&f,&v );
12     for ( int i = 0; i != f; ++i )
13         for ( int j = 0; j != v; ++j )
14             scanf ( "%d",&s1[i][j] );
15     int s2[100][100];
16     int j2 = 0;
17     for ( int i = 0; i != f; ++i ) // 把需要判断的数挑选出来
18     {
19         int k1 = 0;
20         j2 = 0;
21         for ( int j = i; j <= v-f+i ; ++j ) //主要是 v-f+1 控制条件
22             s2[i][j2++] = s1[i][j];
23     }
24     int max[1000];
25     for ( int i = f - 1,j = 0; j != j2;++j )
26         max[j] = s2[i][j];
27     //for ( int j = 0; j != j2; ++j  )
28     //    printf ( "%d \n",max[j] );
29     for ( int i = f - 2; i >= 0; --i ) // DP 主要从最后一行往上判断每个数sum的max;
30     {
31         int max1;
32         for ( int j = 0; j != j2; ++j )
33         {
34             max1 = -500;
35             int sum;
36             for ( int j1 = j; j1 != j2; ++j1 )
37             {
38                 sum = s2[i][j] + max[j1];
39                 if ( sum > max1 )
40                     max1 = sum;
41             }
42             max[j] = max1;
43         }
44     }
45     sort( max,max+j2 );
46     printf ( "%d\n",max[j2-1] );
47     return 0;
48 }
49

haozi 2010-03-23 16:28 发表评论

pku 1050

haozi — Sun, 21 Mar 2010 13:31:00 GMT

1 首先学到的是求一维数组最大子段和! 这个的想法很简单,只要大家多动几次笔,想想就会理解了!
2 我卡住的地方是,为什么如果 <= 0 就从新开始然道 <= 0 这个子段中任意数字重新开始新的子段就没有最大的?
3 最后想想就明白了!
4
5 int p( int *s2,int c1 )
6 {
7  int max = -1000;
8  int sum = 0;
9  for ( int i = 0; i != c1; ++i ) // 一维数组最大子段和
10  {
11   if ( sum > 0 )
12    sum += s2[i];
13   else sum = s2[i];
14   if ( sum > max )
15    max = sum;
16  }
17 }
18
19 然后做了到题目,求矩阵的最大子矩阵的和! 它的思想和一维是一样的!
20
21 #include <cstdio>
22
23 int p( int *s2,int c1 )
24 {
25  int max = -1000;
26  int sum = 0;
27  for ( int i = 0; i != c1; ++i ) // 一维数组最大子段和
28  {
29   if ( sum > 0 )
30    sum += s2[i];
31   else sum = s2[i];
32   if ( sum > max )
33    max = sum;
34  }
35  return max;
36 }
37
38 int main()
39 {
40  int r,c;
41  scanf ( "%d",&r );
42  int s1[500][500];
43  for ( int i = 0; i != r; ++i )
44   for ( int j = 0; j != r; ++j )
45    scanf ( "%d\n",&s1[i][j] );
46  int b[500];
47  int max = -1000;
48  int max1;
49  for ( int i = 0; i != r; ++i ) //控制起始行
50  {
51   for ( int b1 = 0; b1 != r; ++b1 )
52    b[b1] = 0;
53   for ( int j = i; j != r; ++j ) // 控制行数从 i 起始到最后一行;
54   {
55    for ( int k = 0; k != r; ++k )
56     b[k] += s1[j][k];
57    max1 = p( b,r );
58    if ( max1 > max )
59     max = max1;
60   }
61  }
62  printf ( "%d\n",max );
63  return 0;
64 }
65
66

haozi 2010-03-21 21:31 发表评论