poj 3264 Balanced Lineup

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

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <algorithm>

 4 #define maxn 5001000

 5 using namespace std;

 6 

 7 int a[maxn],x,y,N,Q,ans,max1,min1;

 8 

 9 struct node

10 {

11     int l;

12     int r;

13     int tall;

14     int shorte;

15 }p[maxn];

16 

17 void build_tree(int i,int l,int r)

18 {

19     p[i].l=l;

20     p[i].r=r;

21     if(l==r)

22     {

23         p[i].tall=a[l];

24         p[i].shorte=a[l];

25         return;

26     }

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

28     build_tree(i+i,l,mid);

29     build_tree(i+i+1,mid+1,r);

30     p[i].tall=max(p[i+i].tall,p[i+i+1].tall);

31     p[i].shorte=min(p[i+i].shorte,p[i+i+1].shorte);

32 }

33 

34 void search1(int i,int l,int r)

35 {

36     if(p[i].l==l&&p[i].r==r)

37     {

38         max1=max(max1,p[i].tall);

39         min1=min(min1,p[i].shorte);

40         return ;

41     }

42     int mid=(p[i].l+p[i].r)/2;

43     if(r<=mid)

44     {

45         search1(i+i,l,r);

46     }

47     else if(l>mid)

48     {

49         search1(i+i+1,l,r);

50     }

51     else

52     {

53         search1(i+i,l,mid);

54         search1(i+i+1,mid+1,r);

55     }

56 }

57 

58 int main()

59 {

60     while(scanf("%d%d",&N,&Q)!=EOF)

61     {

62         for(int i=1; i<=N; i++)

63         {

64             scanf("%d",&a[i]);

65         }

66         build_tree(1,1,N);

67         for(int i=0; i<Q; i++)

68         {

69             max1=-1;

70             min1=maxn;

71             scanf("%d%d",&x,&y);

72             search1(1,x,y);

73             printf("%d\n",max1-min1);

74         }

75     }

76     return 0;

77 }

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 用动态规划解决的；

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <algorithm>

 4 #include <cmath>

 5 #define maxn 500010

 6 using namespace std;

 7 

 8 int maxsum[maxn][30],a[maxn],N,Q,minsum[maxn][30],x,y;

 9 

10 int main()

11 {

12    while(scanf("%d%d",&N,&Q)!=EOF)

13    {

14        for(int i=1; i<=N; i++)

15        {

16            scanf("%d",&a[i]);

17        }

18        for(int i=1; i<=N; i++)

19        {

20            maxsum[i][0]=a[i];

21            minsum[i][0]=a[i];

22        }

23        int m=(int)(log(N*1.0)/log(2.0));

24        for(int j=1; j<=m; j++)

25        {

26            for(int i=1; i<=N; i++)

27            {

28                if(i+(1<<j)-1<=N)

29                {

30                    maxsum[i][j]=max(maxsum[i][j-1],maxsum[i+(1<<(j-1))][j-1]);

31                    minsum[i][j]=min(minsum[i][j-1],minsum[i+(1<<(j-1))][j-1]);

32                }

33            }

34        }

35        for(int i=0; i<Q; i++)

36        {

37            scanf("%d%d",&x,&y);

38            int k=(int)(log(y-x+1.0)/(log(2.0)));

39            printf("%d\n",max(maxsum[x][k],maxsum[y-(1<<k)+1][k])-min(minsum[x][k],minsum[y-(1<<k)+1][k]));

40        }

41    }

42    return 0;

43 }

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