C++博客-gzwzm06-文章分类-数据结构

7th GDCPC Problem H(DFS序 + 线段树)

Tue, 19 May 2009 16:13:00 GMT

摘要: 1#include 2 3const int SIZE = 100001; 4const int INF = 10000000; 5 6int&n... 阅读全文

巫 2009-05-20 00:13 发表评论

POJ 3368 (RMQ)

Fri, 08 May 2009 00:00:00 GMT

摘要: 1#include 2#include 3 4const int SIZE = 100101; 5 6int arr[SIZE],&nbs... 阅读全文

巫 2009-05-08 08:00 发表评论

POJ 3368 Frequent values (线段树)

Thu, 07 May 2009 23:59:00 GMT

摘要: 1#include 2 3const int SIZE = 100101; 4 5struct STREE 6{ 7 &... 阅读全文

巫 2009-05-08 07:59 发表评论

POJ 2060--Taxi Cab Scheme(最小路径覆盖)

Wed, 22 Apr 2009 12:33:00 GMT

摘要: 最小路径覆盖 = |N| - 最大匹配数 1#include 2#include 3#include 4 5const int SI... 阅读全文

巫 2009-04-22 20:33 发表评论

Atomic Nucleus Investigation (线段树 n = 100000)

Mon, 20 Apr 2009 13:20:00 GMT

摘要: 第一次写接近5KB的代码，故发帖留念.- generate M：如果集合原来没有M，则添加M；否则，不操作- remove M：如果集合存在M，则删除M；否则，不操作- query：查出集合各元素之间最小间隔其中，M （1， 100000） 1#include 2#include 阅读全文

巫 2009-04-20 21:20 发表评论

POJ 2481(树状数组)

Sat, 11 Apr 2009 08:50:00 GMT

1#include <cstdio>
2#include <algorithm>
3using namespace std;
4
5const int SIZE = 100005;
6
7struct RANGE
8{
9    int m_S, m_E;
10    int m_p;
11
12    bool operator < (const RANGE& other) const
13    {
14        if ( m_E != other.m_E )
15            return (m_E > other.m_E);
16
17        return (m_S < other.m_S);
18    }
19}cow[SIZE];
20
21int N, result[SIZE], C[SIZE], max_N;
22
23int LowBit( const int & k )
24{
25    return (k & (-k));
26}
27
28void Modify(int num, int value)
29{
30    while ( num <= max_N )
31    {
32        C[num] += value;
33        num += LowBit(num);
34    }
35}
36
37int GetSum(int num)
38{
39    int sum = 0;
40    while ( num > 0 )
41    {
42        sum += C[num];
43        num -= LowBit(num);
44    }
45
46    return sum;
47}
48
49void Init()
50{
51    for ( int i = 0; i <= max_N; ++i )
52    {
53        C[i] = 0;
54    }
55}
56
57int main()
58{
59    int i;
60
61//    freopen("1.txt", "r", stdin);
62    while ( scanf("%d", &N), N != 0 )
63    {
64        max_N = 0;
65        for ( i = 0; i < N; ++i )
66        {
67            scanf("%d %d", &cow[i].m_S, &cow[i].m_E);
68            cow[i].m_S++, cow[i].m_E++;
69            cow[i].m_p = i;
70
71            if ( max_N < cow[i].m_E )
72                max_N = cow[i].m_E;
73        }
74        Init();
75
76        sort(cow, cow + N);
77
78        for ( i = 0; i < N; ++i )
79        {
80            if ( i != 0 && cow[i].m_S == cow[i - 1].m_S && cow[i].m_E == cow[i - 1].m_E )
81                result[cow[i].m_p] = result[cow[i - 1].m_p];
82            else
83                result[cow[i].m_p] = GetSum( cow[i].m_S );
84
85            Modify( cow[i].m_S, 1 );
86        }
87
88        for ( i = 0; i < N - 1; ++i )
89            printf("%d ", result[i]);
90        printf("%d\n", result[i]);
91    }
92    return 0;
93}

巫 2009-04-11 16:50 发表评论

POJ 2481 Cows(线段树)

Sat, 11 Apr 2009 08:15:00 GMT

摘要: 1#include 2#include 3using namespace std; 4 5const int SIZE = 100001;&n... 阅读全文

巫 2009-04-11 16:15 发表评论

POJ 3667 Hotel(模拟)

Sun, 22 Mar 2009 07:58:00 GMT

摘要: 两种方法：直接模拟、线段树由于编码菜，程序一大堆bug，下载测试数据，调试了两个钟头，才搞太无语了。。。 1#include 2 3const int SIZE = 50001; 4 5struct&nbs... 阅读全文

巫 2009-03-22 15:58 发表评论

Pku 2828(线段树)

Tue, 18 Nov 2008 02:12:00 GMT

利用线段树倒推

1#include <cstdio>
2
3const int MAXN = 200100 ;
4
5struct NODE
6{
7    int start , end ;
8    int sum ;
9    NODE *leftc ;
10    NODE *rightc ;
11
12    void Build( const int& s, const int& e ) ;
13    void Insert( int pos ) ;
14} ;
15
16NODE *root , SegTree[MAXN * 2] ;
17int g_Pos , mark , array[MAXN][2] , seq[MAXN] ;
18
19void NODE::Build( const int& s, const int& e )
20{
21    int mid = ( s + e ) >> 1 ;
22    start = s , end = e ;
23    sum = mid - s + 1 ;    //记录此点左边有多少个人
24
25    if ( s == e )
26    {
27        leftc = rightc = NULL ;
28        return ;
29    }
30
31    leftc = &SegTree[g_Pos++] ;
32    rightc = &SegTree[g_Pos++] ;
33
34    leftc->Build( s , mid ) ;
35    rightc->Build( mid + 1 , e ) ;
36}
37
38void NODE::Insert( int pos )
39{
40    if ( start == end )
41    {
42        mark = start ; //记录最终位置
43        return ;
44    }
45    if ( sum < pos )    //左边人数比指定位置小，则插到右孩子节点
46    {
47        pos -= sum ;
48        rightc->Insert( pos ) ;
49    }
50    else if ( sum >= pos )    //否则插到左孩子节点
51    {
52        sum-- ;        //那么左边的人数空位减1
53        leftc->Insert( pos ) ;
54    }
55}
56
57void Init()
58{
59    root = &SegTree[0] ;
60    g_Pos = 1 ;
61}
62
63int main()
64{
65//    freopen("in.txt", "r", stdin) ;
66
67    int n , i ;
68
69    while ( scanf("%d", &n) != EOF )
70    {
71        Init() ;
72
73        for ( i = 0 ; i < n ; ++i )
74        {
75            scanf("%d %d", &array[i][0], &array[i][1]) ;
76        }
77
78        root->Build( 1, n ) ;
79
80        for ( i = n - 1 ; i >= 0 ; --i )
81        {
82            root->Insert( array[i][0] + 1 ) ;
83
84            seq[mark] = array[i][1] ;
85        }
86
87        for ( i = 1 ; i < n ; ++i )
88            printf("%d ", seq[i]) ;
89        printf("%d\n", seq[i]) ;
90    }
91    return 0 ;
92}

巫 2008-11-18 10:12 发表评论

Pku 2777(线段树)

Mon, 17 Nov 2008 14:55:00 GMT

摘要: 1#include 2#include 3 4const long max = 200001 ; 5 6struct N... 阅读全文

巫 2008-11-17 22:55 发表评论

POJ 2528(离散化 + 线段树)

Sun, 16 Nov 2008 12:58:00 GMT

摘要: 1#include 2#include 3#include 4using namespace std ; 5 &nbs... 阅读全文

巫 2008-11-16 20:58 发表评论

Poj 2299--归并排序

Wed, 12 Nov 2008 16:37:00 GMT

1 #include <stdio.h>
2
3 const int MAXN = 500001 ;
4 __int64 a[MAXN] , c[MAXN] , cnt ;
5
6 void Merge( int le, int mid, int rh);
7
8 void MergeSort(int le, int rh){
9
10
11     if (rh>le)
12     {
13         int mid = (le+rh) >> 1;
14         MergeSort(le,mid);
15         MergeSort(mid + 1,rh);
16         Merge(le, mid, rh);
17     }
18 }
19
20 void Merge(int le, int mid, int rh)
21 {
22         int i, j,
23         tmp = 1;
24         for (i = le, j = mid+1; i <= mid && j <= rh; )
25         {
26             if (a[j] < a[i])
27             {
28                 c[tmp++] = a[j++];
29                 cnt += mid - i + 1;    //记录逆序数
30             }
31             else   c[tmp++] = a[i++];
32         }
33        while ( j <= rh )
34                 c[tmp++] = a[j++];
35        while( i <= mid )
36                 c[tmp++] = a[i++];
37
38         for (i = le; i <= rh; ++i)
39         {
40             a[i] = c[i - le + 1];
41         }
42
43 }
44
45
46 int main()
47 {
48     int n , i ;
49
50     while ( scanf("%d", &n) && n != 0 )
51     {
52         for ( i = 0 ; i < n ; i++ )
53         {
54             scanf("%I64d", &a[i]) ;
55         }
56         cnt = 0 ;
57         MergeSort( 0, n - 1 ) ;
58
59         printf("%I64d\n", cnt) ;
60     }
61     return 0 ;
62 }

巫 2008-11-13 00:37 发表评论