题意：http://acm.hdu.edu.cn/showproblem.php?pid=1506看图一目了然。两个方向单调队列维护下。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <map> 7 #include <queue> 8 #include <deque> 9 #include <cmath> 10 #include <vector> 11 #include <ctime> 12 #include <cctype> 13 #include <set> 14 15 using namespace std; 16 17 #define mem0(a) memset(a, 0, sizeof(a)) 18 #define lson l, m, rt << 1 19 #define rson m + 1, r, rt << 1 | 1 20 #define define_m int m = (l + r) >> 1 21 #define Rep(a, b) for(int a = 0; a < b; a++) 22 #define lowbit(x) ((x) & (-(x))) 23 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 24 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 25 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 26 27 typedef double db; 28 typedef long long LL; 29 typedef pair<int, int> pii; 30 typedef multiset<int> msi; 31 typedef multiset<int>::iterator msii; 32 typedef set<int> si; 33 typedef set<int>::iterator sii; 34 35 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}; 36 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1}; 37 const int maxn = 1e6 + 7; 38 const int maxm = 1e5 + 7; 39 const int MD = 1e9 +7; 40 const int INF = 1e9 + 7; 41 42 template<class T> struct MonotoneQueue{ 43 deque<T> Q; 44 MonotoneQueue<T>() { Q.clear(); } 45 void clear() { Q.clear(); } 46 bool empty() { return Q.empty(); } 47 void add_back(T x) { while (!Q.empty() && !(Q.back() < x)) Q.pop_back(); Q.push_back(x); } 48 void pop_front() { Q.pop_front(); } 49 T back2() { return *(Q.end() - 2); } 50 T front() { return Q.front(); } 51 }; 52 53 struct pair1 { 54 int val, pos; 55 bool operator < (const pair1 &a) const { 56 return val < a.val; 57 } 58 constructInt2(pair1, val, pos); 59 }; 60 61 MonotoneQueue<pair1> UQ; 62 63 int a[100010], L[100010]; 64 65 int main() { 66 //freopen("in.txt", "r", stdin); 67 int n; 68 while (cin >> n, n) { 69 UQ.clear(); 70 UQ.add_back(pair1(-1, -1)); 71 for (int i = 0; i < n; i++) { 72 scanf("%d", a + i); 73 UQ.add_back(pair1(a[i], i)); 74 L[i] = UQ.back2().pos + 1; 75 } 76 UQ.clear(); 77 UQ.add_back(pair1(-1, n)); 78 LL ans = 0; 79 for (int i = n - 1; i >= 0; i--) { 80 UQ.add_back(pair1(a[i], i)); 81 ans = max(ans, (LL)a[i] * (UQ.back2().pos - L[i])); 82 } 83 cout << ans << endl; 84 } 85 return 0; 86 }