HDU1506 Largest Rectangle in a Histogram(算竞进阶习题)

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;
}