HDU 1506 Largest Rectangle in a Histogram

这个问题姑且也叫做最大子矩阵吧

给一个树状图，求一个最大面积的子矩阵

 

思路是这样的，对于每个单位矩阵，求出左边连续不比它低的矩阵的下标，放在l数组里

同样，再求出右边连续的不比它低的矩阵的下标

这样，对于每个单个矩阵所能得到的最大面积就是 (r[i]-l[i]+1)*a[i]

//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 100000 + 5;
long long a[maxn], l[maxn], r[maxn];

int main(void)
{
    #ifdef LOCAL
        freopen("1506in.txt", "r", stdin);
    #endif

    long long n;
    while(scanf("%I64d", &n) == 1 && n)
    {
        long long ans = -1;
        long long i, t;
        for(i = 1; i <= n; ++i)
            scanf("%I64d", &a[i]);
        l[1] = 1;
        r[n] = n;
        for(i = 2; i <= n; ++i)
        {
            t = i;
            while(t > 1 && a[i] <= a[t-1])
                t = l[t-1];
            l[i] = t;
        }
        for(i = n-1; i > 0; --i)
        {
            t = i;
            while(t < n && a[i] <= a[t+1])
                t = r[t+1];
            r[i] = t;
        }
        long long temp;
        for(i = 1; i <= n; ++i)
        {
            temp = (r[i]-l[i]+1)*a[i];
            ans = max(ans, temp);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}