单调栈的运用

最新推荐文章于 2021-07-14 13:36:11 发布

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本文介绍了一种使用单调栈解决寻找直方图中最大矩形面积问题的算法，并提供了完整的AC代码实现。

E - E

Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu

Submit Status

Description

Given n positive integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height =[2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

Input

　　The first line of the input file contains an integer T (T<=100) specifying the number of test cases.
Each test case begins with a line containing an integer N (1<=N<=100000),The following N-1 lines each contain a integer A[i](1<=A[i]<=10000).

Output

For each query, output a single line containing largest area.

Sample Input

Sample Output

10
30

题解： 单调栈

AC代码：

#include <iostream>
#include <stdio.h>
#include <stack>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<int,int> P;
const int M = 1e5 + 5;
stack<P> st;
int a[M],L[M],R[M],N;

void solve(int c[],int val) {
    while(st.size()) st.pop();
    for(int i = 1; i <= N + 1; i++) {
        int x = a[i];
        while(!st.empty() && st.top().first > x) {
            if(val) c[st.top().second] = (i - st.top().second) * st.top().first;
            else c[N - st.top().second + 1] = (i - st.top().second) * st.top().first;
            st.pop();
        }
        st.push(make_pair(x,i));
    }
}

int main() {
    int T;
    cin>>T;
    while(T--) {
        while(st.size() > 0) st.pop();
        int ans = 0;
        scanf("%d",&N);
        for(int i = 1; i <= N; i++) {
            scanf("%d",&a[i]);
        }
        a[N + 1] = 0;
        solve(R,1);
        reverse(a + 1,a + N + 1);
        solve(L,0);
        reverse(a + 1,a + N + 1);
        for(int i = 1; i <= N; i++) ans = max(ans,L[i] + R[i] - a[i]);
        printf("%d\n",ans);
    }
    return 0;
}