子串的最大差

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#436. 子串的最大差

题意

给你一个数组,求所有区间的(最大值-最小值)的和。

思路

对于每个数计算它的贡献。贡献分为两部分,一部分是作为最大值,一部分是作为最小值。当a[i]作为最大值,你要找到它左边第一个大于等于(可能存在连续相等的数,避免重复左闭右开)他的数位置l和右边第一个比他大的数位置r,那么当前树的贡献是a[i] * (i-l)* (r-i)。最小值同理不过是减去。

这里介绍两种寻找每个数的l,r的方法。

  1. 单调栈
    好像就相当是个板子了
  2. 二分+st表
    对于每个数a[i]二分它的左边【0,i-1】和右边[i+1,N],然后用st表来check。一段区间的前缀或后缀的最值是单调的,满足二分。

单调栈代码

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

#define INF 0x3f3f3f3f3f3f3f3f
#define int long long


typedef double db;
typedef long long ll;
typedef unsigned long long ull;

int qmi(int a, int k, int p)int res = 1;while (k)if (k & 1) res = (ll)res * a % p;a = (ll)a * a % p;k >>= 1;return res;
int qpow(int a,int b)int res = 1;while(b)if(b&1) res *= a;b>>=1;a*=a;return res;
int mo(int x,int p)return x = ((x%p)+p)%p;
int gcd(int a,int b)return b?gcd(b,a%b):a;


const int maxn = 1e6+7;
const int mod = 1e9+7;
const double eps = 1e-6;
int dx[] = 0,0,1,-1, dy[] = 1,-1,0,0;

int T = 1,N,M,K,Q;

void solve()
	cin >> N;
	vector<int> a(N+2);
	for (int i = 1; i <= N; i ++) cin >> a[i];
	a[0] = a[N+1] = 1e9;//设置哨兵,
		
	vector<int> maxl(N+2),maxr(N+2),minl(N+2),minr(N+2);
	stack<pair<int, int>> q;//pair是值和位置
	
	q.push(make_pair(a[0], 0));
	for (int i = 1; i <= N + 1; i ++) //i is max
		while (q.size() && q.top().first < a[i]) 
			maxr[q.top().second] = i;//a[i]作为栈顶右边的最大值
			q.pop();
		
		if (q.size()) maxl[i] = q.top().second;//栈顶作为a[i]左边的最大值
		q.push(make_pair(a[i], i));//a[i]入栈
	
	

	while(q.size()) q.pop();//清空栈内元素	
	a[0] = a[N+1] = -1e9;//哨兵
	q.push(make_pair(a[0], 0));
	//求最小值同理
	for (int i = 1; i <= N + 1; i ++) //i is min
		while (q.size() && q.top().first > a[i]) 
			minr[q.top().second] = i;
			q.pop();
		
		if (q.size()) minl[i] = q.top().second;
		q.push(make_pair(a[i], i));
	
	
	int ans = 0;
	for (int i = 1; i <= N; i ++) //枚举每个位置的贡献
		int r = maxr[i] - i;
		int l = i - maxl[i];
		int cnt = r * l;
		r = minr[i] - i;
		l = i - minl[i];
		ans += a[i] * cnt;
	
	cout << ans << endl;


signed main()

	ios::sync_with_stdio(false);cin.tie(0);
	
	// cin >> T;
	for (int i = 1; i <= T; i ++) solve();
    return (0-0); //<3
 

二分+st表代码

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

#define INF 0x3f3f3f3f3f3f3f3f
#define int long long


typedef double db;
typedef long long ll;
typedef unsigned long long ull;

int qmi(int a, int k, int p)int res = 1;while (k)if (k & 1) res = (ll)res * a % p;a = (ll)a * a % p;k >>= 1;return res;
int qpow(int a,int b)int res = 1;while(b)if(b&1) res *= a;b>>=1;a*=a;return res;
int mo(int x,int p)return x = ((x%p)+p)%p;
int gcd(int a,int b)return b?gcd(b,a%b):a;


const int maxn = 1e6+7;
const int mod = 1e9+7;
const double eps = 1e-6;
int dx[] = 0,0,1,-1, dy[] = 1,-1,0,0;

int T = 1,N,M,K,Q;
vector<int> A(maxn), dp(maxn), f(maxn), B(maxn);
int a, b, c, x, t;

template<typename T>
struct RMQ 
    static int highest_bit(unsigned x) 
        return x == 0 ? -1 : 31 - __builtin_clz(x);
    
 
    int n = 0;
    vector<vector<T>> range_max;
 
    RMQ(const vector<T> &values = ) 
        if (!values.empty())
            build(values);
    
 
    static T better(T a, T b) 
        return max(a, b);
    
 
    void build(const vector<T> &values) 
        n = (int)(values.size());
        int levels = highest_bit(n) + 1;
        range_max.resize(levels);
 
        for (int k = 0; k < levels; k++)
            range_max[k].resize(n - (1 << k) + 1);
 
        if (n > 0)
            range_max[0] = values;
 
        for (int k = 1; k < levels; k++)
            for (int i = 0; i <= n - (1 << k); i++)
                range_max[k][i] = better(range_max[k - 1][i], range_max[k - 1][i + (1 << (k - 1))]);
    
 
    T query_value(int a, int b) const //[a,b),idx from 1
        // assert(0 <= a && a < b && b <= n);
        int level = highest_bit(b - a + 1);
        return better(range_max[level][a], range_max[level][b - (1 << level) + 1]);
    
;

template<typename T>
struct RMQ1 
    static int highest_bit(unsigned x) 
        return x == 0 ? -1 : 31 - __builtin_clz(x);
    
 
    int n = 0;
    vector<vector<T>> range_min;
 
    RMQ1(const vector<T> &values = ) 
        if (!values.empty())
            build(values);
    
 
    static T better(T a, T b) 
        return min(a, b);
    
 
    void build(const vector<T> &values) 
        n = (int)(values.size());
        int levels = highest_bit(n) + 1;
        range_min.resize(levels);
 
        for (int k = 0; k < levels; k++)
            range_min[k].resize(n - (1 << k) + 1);
 
        if (n > 0)
            range_min[0] = values;
 
        for (int k = 1; k < levels; k++)
            for (int i = 0; i <= n - (1 << k); i++)
                range_min[k][i] = better(range_min[k - 1][i], range_min[k - 1][i + (1 << (k - 1))]);
    
 
    T query_value(int a, int b) const //[a,b),idx from 1
        // assert(0 <= a && a < b && b <= n);
        int level = highest_bit(b - a + 1);
        return better(range_min[level][a], range_min[level][b - (1 << level) + 1]);
    
;




void solve()
	
	cin >> N;
	
	vector<int> a(N+2), sum(N+1);
	for (int i = 1; i <= N; i ++) cin >> a[i];
	a[0] = a[N+1] = 1e9;
	RMQ<int> rmq(a);//维护区间最大
	a[0] = a[N+1] = -1;
	RMQ1<int> rmq1(a);//维护区间最小
	int ans = 0;
	for (int i = 1; i <= N; i ++) //max
		
		//二分求右边界
		int l = i + 1, r = N+1;
		while (l < r) 
			int mid = (l + r) >> 1;
			if (rmq.query_value(i + 1,mid) > a[i]) r = mid ;
			else 
				l = mid + 1;
			
		
		
		int right = l - i;//暂存右边界的长度
		
		
		//二分左边界
		l = 0, r = i - 1;
		while (l < r) 
			int mid = (l + r + 1) >> 1;
			if (rmq.query_value(mid, i - 1) < a[i]) r = mid - 1;
			else 
				l = mid;
			
		
		ans += right * (i-r) * a[i];//计算贡献
		
	
	

	for (int i = 1; i <= N; i ++) //min同理
		
		int l = i + 1, r = N+1;
		while (l < r) 
			int mid = (l + r) >> 1;
			if (rmq1.query_value(i + 1,mid) < a[i]) r = mid ;
			else 
				l = mid 代码源#436子串的最大差

#436. 子串的最大差(单调栈)

代码源 Div1#103子串的最大差 Codeforces - 817D,力扣2104,1900分

代码源 Div1#103子串的最大差 Codeforces - 817D,力扣2104,1900分

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