RMQ算法 

 题目链接：
 http://acm.nyist.net/JudgeOnline/problem.php?pid=119
 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstdio>
 4 using namespace std;
 5 #define maxn 100001
 6 
 7 int ma[maxn][17], mi[maxn][17], N, Q;
 8 //状态转移方程为：
 9 //dp[j][i] = max(dp[j][i-1], dp[j+(1<<(i-1))][i-1]);
10 void rmq_init(){
11     int i, j, t;
12     int m = int(log(double(N))/log(2.0));
13     for(i = 1; i <= m; ++i)
14         for(j = 1; j <= N; ++j){
15             t = j + (1 << (i - 1));
16             if(t <= N){
17                 ma[j][i] = max(ma[j][i-1], ma[t][i-1]);
18                 mi[j][i] = min(mi[j][i-1], mi[t][i-1]);
19             }
20             else{
21                 ma[j][i] = ma[j][i-1];
22                 mi[j][i] = mi[j][i-1];
23             }
24         }
25 }
26 
27 int query(int m, int n)
28 {
29     int t = int(log(double(n - m + 1))/log(2.0));
30     int a = max(ma[m][t], ma[n - (1<<t) + 1][t]);  //区间最大值
31     int b = min(mi[m][t], mi[n - (1<<t) + 1][t]); //区间最小值
32     return a - b;
33 }
34 
35 int main()
36 {
37     int i, m, n;
38     while(scanf("%d%d",&N, &Q) != EOF){
39         for(i = 1; i <= N; ++i){      //初值(dp[i][0])
40             scanf("%d",&ma[i][0]);
41             mi[i][0] = ma[i][0];
42         }
43         rmq_init();
44         for(i = 0; i < Q; ++i)
45         {
46             scanf("%d%d",&m, &n);
47             printf("%d\n",query(m, n));
48         }
49     }
50     return 0;
51 }

 

posted @ 2013-07-09 17:44  YaLing  阅读(377)  评论(0)    收藏  举报