[HDOJ5289]Assignment（RMQ，二分）

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。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 typedef pair<int, int> pii;
 const int maxn = ;
 int dp[maxn][][];
 int a[maxn];
 int n, k;

 void st(int* a, int* b, int n) {
     for(int i = ; i <= n; i++) dp[i][][] = b[i], dp[i][][] = a[i];
     for(int j = ; ( << j) -  <= n; j++) {
         for(int i = ; i + ( << j) -  <= n; i++) {
             dp[i][j][] = min(dp[i][j-][], dp[i+(<<(j-))][j-][]);
             dp[i][j][] = max(dp[i][j-][], dp[i+(<<(j-))][j-][]);
         }
     }
 }

 pii query(int l, int r) {
     int k = int(log(r-l+) / log(2.0));
     return pii(min(dp[l][k][], dp[r-(<<k)+][k][]), max(dp[l][k][], dp[r-(<<k)+][k][]));
 }

 int main() {
     // freopen("in", "r", stdin);
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%d%d",&n,&k);
         for(int i = ; i <= n; i++) scanf("%d", &a[i]);
         st(a, a, n);
         LL ret = ;
         for(int i = ; i <= n; i++) {
             int lo = i, hi = n;
             while(lo <= hi) {
                 int mid = (lo + hi) >> ;
                 pii q = query(i, mid);
                 int minn = q.first, maxx = q.second;
                 if(maxx - minn < k) lo = mid + ;
                 else hi = mid - ;
             }
             ret += (hi - i + );
         }
         printf("%I64d\n", ret);
     }
     return ;
 }