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]

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 using namespace std;

 6 

 7 const int maxn = 100000 + 5;

 8 long long a[maxn], l[maxn], r[maxn];

 9 

10 int main(void)

11 {

12     #ifdef LOCAL

13         freopen("1506in.txt", "r", stdin);

14     #endif

15 

16     long long n;

17     while(scanf("%I64d", &n) == 1 && n)

18     {

19         long long ans = -1;

20         long long i, t;

21         for(i = 1; i <= n; ++i)

22             scanf("%I64d", &a[i]);

23         l[1] = 1;

24         r[n] = n;

25         for(i = 2; i <= n; ++i)

26         {

27             t = i;

28             while(t > 1 && a[i] <= a[t-1])

29                 t = l[t-1];

30             l[i] = t;

31         }

32         for(i = n-1; i > 0; --i)

33         {

34             t = i;

35             while(t < n && a[i] <= a[t+1])

36                 t = r[t+1];

37             r[i] = t;

38         }

39         long long temp;

40         for(i = 1; i <= n; ++i)

41         {

42             temp = (r[i]-l[i]+1)*a[i];

43             ans = max(ans, temp);

44         }

45         printf("%I64d\n", ans);

46     }

47     return 0;

48 }

代码君