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 }
View Code

 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 }
View Code

 

posted @ 2014-02-19 09:41  null1019  阅读(123)  评论(0)  编辑  收藏  举报