poj3264Balanced Lineup(线段树RMQ）

RMQ (Range Minimum/Maximum Query)问题是指：对于长度为n的数列A，回答若干询问RMQ(A,i,j)(i,j<=n)，返回数列A中下标在i,j里的最小(大）值，也就是说，RMQ问题是指求区间最值的问题。

http://poj.org/problem?id=3264

1A 程序跑的好慢 3000+

输完更新 父节点的最小最大值 找的时候找两次 一次最大 一次最小 相减

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 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 #define N 50001

 6 #define MAX 1000001

 7 int s[N*4],mi[N*4],ma[N*4];

 8 void upset(int l,int r,int w)

 9 {

10     if(l==r)

11     return ;

12     int m = (l+r)/2;

13     upset(l,m,2*w);

14     upset(m+1,r,2*w+1);

15     mi[w] = min(mi[w*2],mi[2*w+1]);

16     ma[w] = max(ma[w*2],ma[2*w+1]);

17 }

18 void build(int l,int r,int w)

19 {

20     if(l==r)

21     {

22         scanf("%d", &s[w]);

23         mi[w] = s[w];

24         ma[w] = s[w];

25         return ;

26     }

27     int m = (l+r)/2;

28     build(l,m,2*w);

29     build(m+1,r,2*w+1);

30 }

31 int search(int a,int b,int l,int r,int w)

32 {

33     if(a==l&&b==r)

34     return mi[w];

35     int m = (l+r)/2,re = MAX;

36     if(b<=m)

37     re = min(re,search(a,b,l,m,2*w));

38     else

39     if(a>m)

40     re = min(re,search(a,b,m+1,r,2*w+1));

41     else

42     {

43         re = min(re,search(a,m,l,m,2*w));

44         re = min(re,search(m+1,b,m+1,r,2*w+1));

45     }

46     return re;

47 }

48 int search1(int a,int b,int l,int r,int w)

49 {

50     if(a==l&&b==r)

51     {

52         return ma[w];

53     }

54     int m = (l+r)/2,re = -1;

55     if(b<=m)

56     re = max(re,search1(a,b,l,m,2*w));

57     else

58     if(a>m)

59     re = max(re,search1(a,b,m+1,r,2*w+1));

60     else

61     {

62         re = max(re,search1(a,m,l,m,2*w));

63         re = max(re,search1(m+1,b,m+1,r,2*w+1));

64     }

65     return re;

66 }

67 int main()

68 {

69     int n,q,i,j,k,a,b;

70     while(scanf("%d%d", &n,&q)!=EOF)

71     {

72         build(1,n,1);

73         upset(1,n,1);

74         while(q--)

75         {

76             scanf("%d%d", &a,&b);

77             printf("%d\n",search1(a,b,1,n,1)-search(a,b,1,n,1));

78         }

79     }

80     return 0;

81 }