单调栈裸题

如果矩形高度从左到右是依次递增，那我们枚举每个矩形高度，宽度拉到最优，计算最大面积即可

当有某个矩形比前一个矩形要矮的时候，这块面积的高度就不能大于他本身，所以之前的所有高于他的矩形多出来的部分都没用了，不会再计算第二次。

因此我们只需要用单调栈维护矩形高度即可，当有高度较低的矩形进栈时，先将之前比他高的每个矩形弹出，宽度累加，更新答案。然后我们要进栈的矩形的宽度也要变成之前累加的宽度+1，因为要保留有用部分。。

有个小技巧：为了方便程序，在末尾加一个高度为0的矩形

#include <bits/stdc++.h>

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

inline int lowbit(int x){ return x & (-x); }

inline int read(){

    int X = 0, w = 0; char ch = 0;

    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }

    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();

    return w ? -X : X;

}

inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }

inline int lcm(int a, int b){ return a / gcd(a, b) * b; }

template<typename T>

inline T max(T x, T y, T z){ return max(max(x, y), z); }

template<typename T>

inline T min(T x, T y, T z){ return min(min(x, y), z); }

template<typename A, typename B, typename C>

inline A fpow(A x, B p, C yql){

    A ans = 1;

    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;

    return ans;

}

ll h[100005], s[100005], w[100005], tot;

int main(){

    int n;

    while(cin >> n && n){

        memset(h, 0, sizeof h);

        memset(s, 0, sizeof s);

        memset(w, 0, sizeof w);

        ll ans = 0; w[n + 1] = 0; tot = 0;

        for(int i = 1; i <= n; i ++) cin >> h[i];

        for(int i = 1; i <= n + 1; i ++){

            if(s[tot] > h[i]){

                ll width = 0;

                while(tot > 0 && s[tot] > h[i]){

                    width += w[tot];

                    ans = max(ans, (ll)width * s[tot]);

                    tot --;

                }

                s[++tot] = h[i], w[tot] = width + 1;

            }

            else{

                s[++tot] = h[i], w[tot] = 1;

            }

        }

        printf("%lld\n", ans);

    }

    return 0;

}

再来个笛卡尔树做法

#include <bits/stdc++.h>

#define INF 0x3f3f3f3f

#define full(a, b) memset(a, b, sizeof a)

#define __fastIn ios::sync_with_stdio(false), cin.tie(0)

#define pb push_back

using namespace std;

typedef long long LL;

inline int lowbit(int x){ return x & (-x); }

inline int read(){

    int ret = 0, w = 0; char ch = 0;

    while(!isdigit(ch)){

        w |= ch == '-', ch = getchar();

    }

    while(isdigit(ch)){

        ret = (ret << 3) + (ret << 1) + (ch ^ 48);

        ch = getchar();

    }

    return w ? -ret : ret;

}

template <typename A>

inline A __lcm(A a, A b){ return a / __gcd(a, b) * b; }

template <typename A, typename B, typename C>

inline A fpow(A x, B p, C lyd){

    A ans = 1;

    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;

    return ans;

}

const int N = 200005;

LL h[N], ans;

int n;

struct Node{

    int i, size;

    LL val;

    Node *fa, *lf, *rf;

    Node(){}

    Node(int i, LL val): i(i), val(val){

        fa = lf = rf = nullptr;

        size = 0;

    }

}*root;

Node *buildTree(){

    stack<Node*> st;

    Node *last = nullptr;

    for(int i = 1; i <= n; i ++){

        Node *cur = new Node(i, h[i]);

        while(!st.empty()){

            if(st.top()->val < cur->val){

                Node *up = st.top();

                if(up->rf) cur->lf = up->rf, up->rf->fa = cur;

                up->rf = cur, cur->fa = up;

                break;

            }

            last = st.top(); st.pop();

        }

        if(st.empty() && last) last->fa = cur, cur->lf = last;

        st.push(cur);

    }

    Node *root = nullptr;

    while(!st.empty()) root = st.top(), st.pop();

    return root;

}

void dfs(Node *rt){

    if(rt == nullptr) return;

    rt->size = 1;

    dfs(rt->lf), dfs(rt->rf);

    if(rt->lf) rt->size += rt->lf->size;

    if(rt->rf) rt->size += rt->rf->size;

    ans = max(ans, 1LL * rt->size * rt->val);

}

int main(){

    while(~scanf("%d", &n) && n){

        for(int i = 1; i <= n; i ++) scanf("%lld", &h[i]);

        root = buildTree();

        ans = 0, dfs(root);

        printf("%lld\n", ans);

    }

    return 0;

}