POJ 3264-Balanced Lineup-RMQ问题

裸RMQ问题

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4
 5 using namespace std;
 6
 7 const int MAXN = 50010;
 8
 9 int min_dp[MAXN][20],max_dp[MAXN][20];
10 int min_mm[MAXN],max_mm[MAXN];
11 int b[MAXN];
12 int N,Q;
13
14 void min_initRMQ(int n,int b[])
15 {
16         min_mm[0] = -1;
17         for(int i=1;i<=n;i++)
18         {
19                 min_mm[i] = ( (i&(i-1)) == 0) ? min_mm[i-1]+1 : min_mm[i-1];
20                 min_dp[i][0] = b[i];
21         }
22         for(int j=1;j<=min_mm[n];j++)
23                 for(int i=1;i + (1<<j) - 1 <= n;i++)
24                         min_dp[i][j] = min(min_dp[i][j-1],min_dp[i+(1<<(j-1))][j-1]);
25
26 }
27
28 int min_rmq(int x,int y)
29 {
30         int k = min_mm[y-x+1];
31         return min(min_dp[x][k],min_dp[y-(1<<k)+1][k]);
32 }
33
34
35 void max_initRMQ(int n,int b[])
36 {
37         max_mm[0] = -1;
38         for(int i=1;i<=n;i++)
39         {
40                 max_mm[i] = ( (i&(i-1)) == 0) ? max_mm[i-1]+1 : max_mm[i-1];
41                 max_dp[i][0] = b[i];
42         }
43         for(int j=1;j<=max_mm[n];j++)
44                 for(int i=1;i + (1<<j) - 1 <= n;i++)
45                         max_dp[i][j] = max(max_dp[i][j-1],max_dp[i+(1<<(j-1))][j-1]);
46
47 }
48
49 int max_rmq(int x,int y)
50 {
51         int k = max_mm[y-x+1];
52         return max(max_dp[x][k],max_dp[y-(1<<k)+1][k]);
53 }
54
55 int main()
56 {
57         //freopen("input.in","r",stdin);
58         while(~scanf("%d%d",&N,&Q))
59         {
60                 for(int i=1;i<=N;i++)
61                 {
62                     scanf("%d",&b[i]);
63                 }
64                 min_initRMQ(N,b);
65                 max_initRMQ(N,b);
66
67                 for(int i=0,l,r;i<Q;i++)
68                 {
69                         scanf("%d%d",&l,&r);
70                         printf("%d\n",max_rmq(l,r) - min_rmq(l,r) );
71                         //printf("%d\n",max_rmq(l,r));
72                 }
73         }
74 }