单调栈简单题目整理

最大矩阵类问题

1.求最大矩阵

题目链接

题目：直方图中的最大矩阵

#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
//#define int long long
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e6+10;
int n;
ll h[N];
int w[N], p[N];
void sol() {
    for(int i = 1;i <= n; ++i) scanf("%lld", &h[i]);
    h[n+1] = 0;
    int c = 0;
    ll ma = 0;
    for(int i = 1;i <= n+1; ++i) {
        ll x = h[i];
        if(h[i]>=p[c]) {
            p[++c] = h[i];
            w[c] = 1;
        }
        else {
            int width = 0;
            while(h[i]<p[c]) {
                width += w[c];
                ma = max(ma, 1ll*width*p[c--]);
            }
            p[++c] = h[i]; w[c] = width+1;
        }
    }
    printf("%lld\n", ma);
}
int main()
{
    while(scanf("%d", &n), n)
    sol();
    return 0;
}

2 求最大正方形，也就是最大矩阵稍微改改

题目链接

题目：城市游戏

需要先预处理从当前位置向右边连续的F的个数O(n^2)
然后枚举起始列，对每一列进行一个单调栈处理出最大正方形O(n^2)

#include<bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<string, int> pii;
typedef pair<int, int> piii;
typedef unsigned long long ull;
const int N = 2010, maxn = 1e6+10;
const ll mod = 1e9+7;

int n, m, k;
char s[N][N];
int num[N][N];

int calc(int x)
{
    int h[N], w[N], p[N], c = 0;
    for(int i = 1; i <= n; ++i) h[i] = num[i][x], w[i] = p[i] = 0;
    h[0] = 0;
    h[n+1] = 0;
    p[0] = 0;

    int ans = 0;
    for(int i = 1;i <= n+1; ++i) {
        if(h[i] >= p[c]) p[++c] = h[i], w[c] = 1;
        else {
            int width = 0;
            while(h[i] < p[c]) {
                width += w[c];
                ans = max(ans, width*p[c--]);
            }
            p[++c] = h[i], w[c] = width+1;
        }
    }
    return ans;
}
void sol() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) scanf(" %c", &s[i][j]);

    for(int i = 1;i <= n; ++i)
    for(int j = 1;j <= m; ++j)
    {
        if(s[i][j] == 'F') num[i][j] = 1;
        else continue;
        if(num[i][j-1]>1&&s[i][j]==s[i][j-1]) {
            num[i][j] = num[i][j-1]-1;
        }
        else {
            int k = j;
            while(k+1 <= m&&s[i][j]==s[i][k+1]) k++, num[i][j]++;
        }
    }
    int ans = 0;
    for(int st = 1; st <= m; ++st)
    ans = max(ans, calc(st));
    printf("%lld\n", 3ll*ans);
}
signed main()
{
    sol();
}

最近的最高或最低的位置问题。

1 模板类

题目链接

题目：仰视奶牛

#include<bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<string, int> pii;
typedef pair<int, int> piii;
typedef unsigned long long ull;
const int N = 2e5+10, maxn = 1e6+10;
const ll mod = 1e9+7;
int n, m, k;
int h[N], p[N], ans[N];
void sol() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &h[i]);
    int c = 0;
    for(int i = 1; i <= n; ++i)
    {
        if(c==0||h[i]<=h[p[c]]) p[++c] = i;
        else {
            while(c!=0&&h[i]>h[p[c]]) {
                ans[p[c--]] = i;
            }
            p[++c] = i;
        }
    }
    for(int i = 1;i <= n; ++i)
    printf("%d\n", ans[i]);
}
signed main()
{
    sol();
}

2 接雨水类

题目链接

题目：接雨水

维护一个单调严格递减的栈，若当前柱子高于栈顶的柱子，退栈直到符合单调性；否则加入栈顶。每次退栈时，累加雨水的体积（每次算一层）

#include<bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define ALL(a) a.begin(), a.end()
#define SZ(a) (int)a.size()
#define db1(x) cout<<#x<<" = "<<x<<endl
#define db2(x, y) cout<<#x<<" = "<<x<<" "<<#y<<" = "<<y<<endl
#define endl "\n"
#define inf 0x3f3f3f3f
#define MEM(a, b) memset(a, b, sizeof(a))
//#define int long long
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef unsigned long long ull;
const int N = 2e5+10, maxn = 1e6+10;
const ll mod = 1e9+7;

int n, m, k;
int h[N];
void sol() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) scanf("%d", &h[i]);
    stack<int> st;
    ll ans = 0;
    for(int i = 1;i <= n; ++i)
    {
        while(!st.empty()&&h[st.top()]<h[i]) {
            int t = st.top(); st.pop();
            if(!st.empty()){
                int l = i-st.top()-1;
                int w = min(h[i], h[st.top()])-h[t];
                ans += l*w;
            }
        }
        st.push(i);
    }
    printf("%lld\n", ans);
}
signed main()
{
    sol();
}

本文地址：https://blog.csdn.net/weixin_44070289/article/details/109255222