poj3468A Simple Problem with Integers(成段操作）

 

 

 

http://acm.sdut.edu.cn/bbs/read.php?tid=5651

。。本来只会向上更新 现在学习了如何向下更新 延迟标记法。。。使复杂度降低

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 1 #include <iostream>

 2 #include<cstdio>

 3 #include<string.h>

 4 using namespace std;

 5 #define max 100000

 6 __int64 s[max*4],te[max*4];

 7 void pushup(int w)

 8 {

 9     s[w] = s[2*w]+s[2*w+1];

10 }

11 void pushdown(int w,int m)

12 {

13     if(te[w])

14     {

15         te[2*w]+=te[w];

16         te[2*w+1] += te[w];

17         s[2*w] += (te[w]*(m-(m/2)));

18         s[2*w+1]+=(te[w]*(m/2));

19         te[w] = 0;//将延迟标记去除

20     }

21 }

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

23 {

24     te[w] = 0;

25     if(l==r)

26     {

27         scanf("%I64d",&s[w]);

28         return ;

29     }

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

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

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

33     pushup(w);

34 }

35 void add(int a,int b,int da,int l,int r,int w)

36 {

37     if(a<=l&&b>=r)

38     {

39         te[w] += da;//多次延迟标记

40         s[w]+=da*(r-l+1);

41         return ;

42     }

43     pushdown(w,r-l+1);//更新

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

45     if(a<=m)

46     add(a,b,da,l,m,2*w);

47     if(b>m)

48     add(a,b,da,m+1,r,2*w+1);

49     pushup(w);

50 }

51 __int64 search(int a,int b,int l,int r,int w)

52 {

53     if(a<=l&&b>=r)

54     {

55         return s[w];

56     }

57     pushdown(w,r-l+1);

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

59     __int64 re = 0;

60     if(a<=m)

61     re+=search(a,b,l,m,2*w);

62     if(b>m)

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

64     return re;

65 }

66 int main()

67 {

68     int i,j,k,n,m,t,a,b,c;

69     char z;

70     scanf("%d%d",&n,&m);

71     build(1,n,1);

72     while(m--)

73     {

74         scanf("%*c%c",&z);

75         if(z=='Q')

76         {

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

78             printf("%I64d\n",search(a,b,1,n,1));

79         }

80         else

81         {

82             scanf("%d%d%d",&a,&b,&c);

83             add(a,b,c,1,n,1);

84         }

85     }

86     return 0;

87 }