[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 ;
}
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