[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 = 200200;
int dp[maxn][20][2];
int a[maxn];
int n, k;

void st(int* a, int* b, int n) {
    for(int i = 1; i <= n; i++) dp[i][0][0] = b[i], dp[i][0][1] = a[i];
    for(int j = 1; (1 << j) - 1 <= n; j++) {
        for(int i = 1; i + (1 << j) - 1 <= n; i++) {
            dp[i][j][0] = min(dp[i][j-1][0], dp[i+(1<<(j-1))][j-1][0]);
            dp[i][j][1] = max(dp[i][j-1][1], dp[i+(1<<(j-1))][j-1][1]);
        }
    }
}

pii query(int l, int r) {
    int k = int(log(r-l+1) / log(2.0));
    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]));
}

int main() {
    // freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d",&n,&k);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        st(a, a, n);
        LL ret = 0;
        for(int i = 1; i <= n; i++) {
            int lo = i, hi = n;
            while(lo <= hi) {
                int mid = (lo + hi) >> 1;
                pii q = query(i, mid);
                int minn = q.first, maxx = q.second;
                if(maxx - minn < k) lo = mid + 1;
                else hi = mid - 1;
            }
            ret += (hi - i + 1);
        }
        printf("%I64d
", ret);
    }
    return 0;
}