Codeforces 689D Friends and Subsequences（二分+RMQ）

题意：

给你两个数列a和b，问你其中有多少对l和r使得maxri=lai=minri=lbi。

这个题有一个很正宗要的性质：假设我们固定l，那么对于l到r这个区间内的maxri=lai−minri=lbi≤maxr+1i=lai−minr+1i=lbi，那么也就是说，这个值是有单调性的，所以对于每一个l，我们可以通过二分查找，找到一个rmin和rmax是的上面的值取等号，那么ans+=rmax−rmin+1。对于求那个最大最小可以通过线段树或者是RMQ，线段树的复杂度要到O(n(logn)2可能会被卡常数，所以还是用RMQO(1)查询比较好。

代码：

//
//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <sstream>
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
#define SI(a) scanf("%d", &a)
#define SII(a,b) scanf("%d%d", &a, &b)
#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7f7f7f7f;
const int MOD=1e9+7;
const double eps=1e-10;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

const int N = 200000 + 100;

int n;
int a
, b
;

struct SparseTable {
int Max[20]
, Min[20]
;
void init(int n) {
for (int i = 1; i <= n; i++) {
Max[0][i] = a[i];
Min[0][i] = b[i];
}
for (int i = 1; (1 << i) <= n; i++) {
for (int j = 1; j + (1 << i) - 1 <= n; j++) {
Max[i][j] = max(Max[i - 1][j], Max[i - 1][j + (1 << (i - 1))]);
Min[i][j] = min(Min[i - 1][j], Min[i - 1][j + (1 << (i - 1))]);
}
}
}
int RMQ(int l, int r) {
int k = 31 - __builtin_clz(r - l + 1);
return max(Max[k][l], Max[k][r - (1 << k) + 1]) - min(Min[k][l], Min[k][r - (1 << k) + 1]);
}
}st;

int getMin(int l) {
int ans = -1;
int L = l, R = n;
while (L <= R) {
int mid = L + R >> 1;
if (st.RMQ(l, mid) >= 0) ans = mid, R = mid - 1;
else L = mid + 1;
}
if (ans == -1) return -1;
if (st.RMQ(l, ans) == 0) return ans;
else return -1;
}

int getMax(int l) {
int ans = -1;
int L = l, R = n;
while (L <= R) {
int mid = L + R >> 1;
if (st.RMQ(l, mid) <= 0) ans = mid, L = mid + 1;
else R = mid - 1;
}
if (ans == -1) return -1;
if (st.RMQ(l, ans) == 0) return ans;
else return -1;
}

int main(int argc, char const *argv[]) {
#ifdef LOCAL
freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
// freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
while(SI(n) == 1) {
for (int i = 1; i <= n; i++) SI(a[i]);
for (int i = 1; i <= n; i++) SI(b[i]);
st.init(n);
ll ans = 0;
for (int l = 1; l <= n; l++) {
int Rmin = getMin(l);
if (Rmin == -1) continue;
int Rmax = getMax(l);
if (Rmax == -1) continue;
ans += Rmax - Rmin + 1;
}
cout << ans << endl;
}
return 0;
}
/*
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(| -_- |)
O\  =  /O
____/`---'\____
.'  \|     |//  `.
/  \|||  :  |||//  \
/  _||||| -:- |||||-  \
|   | \\  -  /// |   |
| \_|  ''\---/''  |   |
\  .-\__  `-`  ___/-. /
___`. .'  /--.--\  `. . __
."" '<  `.___\_<|>_/___.'  >'"".
| | :  `- \`.;`\ _ /`;.`/ - ` : | |
\  \ `-.   \_ __\ /__ _/   .-` /  /
======`-.____`-.___\_____/___.-`____.-'======
`=---='
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
佛祖保佑        永无BUG
*/