hdu 4638 Group

理解题意：连续的数分为一组，每次判断下就可以了！！

代码如下：

 

 1 #include<iostream>

 2 #include<stdio.h>

 3 #include<algorithm>

 4 #include<iomanip>

 5 #include<cmath>

 6 #include<cstring>

 7 #include<vector>

 8 #define ll __int64

 9 #define pi acos(-1.0)

10 #define MAX 100105

11 using namespace std;

12 int R,L,an[MAX],ans[MAX],res;

13 bool vis[MAX];

14 struct node

15 {

16     int x,y,l,p;

17 }q[MAX];

18 bool cmp(const node &a,const node &b)

19 {

20     if(a.l==b.l) return a.y<b.y;

21     return a.l<b.l;

22 }

23 void query(int x,int y,int flag)

24 {

25     if(flag){

26         for(int i=x;i<L;i++){

27             vis[an[i]]=1;

28             if(vis[an[i]-1]&&vis[an[i]+1]) res--;

29             else if(!vis[an[i]-1]&&!vis[an[i]+1]) res++;

30         }

31         for(int i=R+1;i<=y;i++){

32             vis[an[i]]=1;

33             if(vis[an[i]-1]&&vis[an[i]+1]) res--;

34             else if(!vis[an[i]-1]&&!vis[an[i]+1]) res++;

35         }

36         for(int i=L;i<x;i++){

37             vis[an[i]]=0;

38             if(vis[an[i]-1]&&vis[an[i]+1]) res++;

39             else if(!vis[an[i]-1]&&!vis[an[i]+1]) res--;

40         }

41         for(int i=y+1;i<=R;i++){

42             vis[an[i]]=0;

43             if(vis[an[i]-1]&&vis[an[i]+1]) res++;

44             else if(!vis[an[i]-1]&&!vis[an[i]+1]) res--;

45         }

46     }

47     else{

48         for(int i=x;i<=y;i++){

49             vis[an[i]]=1;

50             if(vis[an[i]-1]&&vis[an[i]+1]) res--;

51             else if(!vis[an[i]-1]&&!vis[an[i]+1]) res++;

52         }

53     }

54     L=x;R=y;

55 }

56 int main(){

57     int n,t,i,j,m,c;

58     scanf("%d",&c);

59     while(c--){

60         scanf("%d%d",&n,&t);

61         for(i=1;i<=n;i++) scanf("%d",&an[i]);

62         m=sqrt(1.0*n);

63         for(i=0;i<t;i++){

64             scanf("%d%d",&q[i].x,&q[i].y);

65             q[i].l=q[i].x/m;

66             q[i].p=i;

67         }

68         sort(q,q+t,cmp);

69         res=0;

70         memset(vis,0,sizeof(vis));

71         for(i=0;i<t;i++){

72             query(q[i].x,q[i].y,i);

73             ans[q[i].p]=res;

74         }

75         for(i=0;i<t;i++)

76             printf("%d\n",ans[i]);

77     }

78     return 0;

79 }

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