hdu1506(dp减少重复计算）
可以算出以第i个值为高度的矩形可以向左延伸left[i]，向右延伸right[i]的长度
那么答案便是 (left[i] + right[i] + 1) * a[i] 的最大值
关键left[i] 和right[i]的计算
如果a[i] > a[i-1] , 那么left[i] = 0
如果a[i] <=a[i-1], 那么left[i] = left[i-1] + 1, 但是位置i-1-left[i-1]-1及其往左的元素没有比较过，所以需要继续去比较
同理right[i]的计算也是一样的
1 #pragma warning(disable:4996) 2 #pragma comment(linker, "/STACK:1024000000,1024000000") 3 #include <stdio.h> 4 #include <string.h> 5 #include <time.h> 6 #include <math.h> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <stack> 11 #include <vector> 12 #include <bitset> 13 #include <algorithm> 14 #include <iostream> 15 #include <string> 16 #include <functional> 17 #include <unordered_map> 18 typedef __int64 LL; 19 const int INF = 999999999; 20 21 const int N = 100000 + 10; 22 int a[N]; 23 LL left[N], right[N]; 24 int main() 25 { 26 int n; 27 while (scanf("%d", &n), n) 28 { 29 for (int i = 1;i <= n;++i) 30 { 31 scanf("%d", &a[i]); 32 } 33 left[1] = right[n] = 0; 34 int l,r; 35 for (int i = 2;i <= n;++i) 36 { 37 l = i - 1; 38 left[i] = 0; 39 while (l>=1 && a[i] <= a[l]) 40 { 41 left[i] += left[l] + 1; 42 l = l - left[l] - 1; 43 } 44 } 45 for (int i = n - 1;i >= 1;--i) 46 { 47 right[i] = 0; 48 r = i + 1; 49 while (r <= n && a[i] <= a[r]) 50 { 51 right[i] += right[r] + 1; 52 r = r + right[r] + 1; 53 } 54 } 55 LL ans = 0; 56 for (int i = 1;i <= n;++i) 57 { 58 ans = std::max(ans, (left[i] + right[i] + 1)*a[i]); 59 } 60 printf("%I64d\n", ans); 61 } 62 return 0; 63 }
