[HDOJ5289]Assignment(RMQ,二分)

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5289

题意:求满足区间内最大值和最小值差为k的区间个数。

预处理出区间的最值,枚举左端点,根据最值的单调性二分枚举右端点,使得找到最右侧max-min<k,区间数为[i,...hi]的个数,即hi-i+1个。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 typedef pair<int, int> pii;
 6 const int maxn = 200200;
 7 int dp[maxn][20][2];
 8 int a[maxn];
 9 int n, k;
10 
11 void st(int* a, int* b, int n) {
12     for(int i = 1; i <= n; i++) dp[i][0][0] = b[i], dp[i][0][1] = a[i];
13     for(int j = 1; (1 << j) - 1 <= n; j++) {
14         for(int i = 1; i + (1 << j) - 1 <= n; i++) {
15             dp[i][j][0] = min(dp[i][j-1][0], dp[i+(1<<(j-1))][j-1][0]);
16             dp[i][j][1] = max(dp[i][j-1][1], dp[i+(1<<(j-1))][j-1][1]);
17         }
18     }
19 }
20 
21 pii query(int l, int r) {
22     int k = int(log(r-l+1) / log(2.0));
23     return pii(min(dp[l][k][0], dp[r-(1<<k)+1][k][0]), max(dp[l][k][1], dp[r-(1<<k)+1][k][1]));
24 }
25 
26 int main() {
27     // freopen("in", "r", stdin);
28     int T;
29     scanf("%d", &T);
30     while(T--) {
31         scanf("%d%d",&n,&k);
32         for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
33         st(a, a, n);
34         LL ret = 0;
35         for(int i = 1; i <= n; i++) {
36             int lo = i, hi = n;
37             while(lo <= hi) {
38                 int mid = (lo + hi) >> 1;
39                 pii q = query(i, mid);
40                 int minn = q.first, maxx = q.second;
41                 if(maxx - minn < k) lo = mid + 1;
42                 else hi = mid - 1;
43             }
44             ret += (hi - i + 1);
45         }
46         printf("%I64d\n", ret);
47     }
48     return 0;
49 }

 

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