挺好的一道題呢

O(n^2)或者O(wh)

1 #include<cstdio> 2 #include<cstring> 3 #include<cstdlib> 4 #include<algorithm> 5 #include<iostream> 6 7 using namespace std; 8 9 void setIO(const string& s) { 10 freopen((s + ".in").c_str(), "r", stdin); 11 freopen((s + ".out").c_str(), "w", stdout); 12 } 13 14 template<typename Q> Q read(Q& x) { 15 static char c, f; 16 for(f = 0; c = getchar(), !isdigit(c); ) if(c == '-') f = 1; 17 for(x = 0; isdigit(c); c = getchar()) x = x * 10 + c - '0'; 18 return f && (x = -x), x; 19 } 20 template<typename Q> Q read() { 21 static Q x; return read(x); 22 } 23 24 const int N = 5000 + 10; 25 26 struct Node { 27 int x, y; 28 Node(int x = 0, int y = 0) : x(x), y(y) {} 29 }p[N]; 30 31 bool cmpx(const Node& lhs, const Node& rhs) { 32 return lhs.x < rhs.x; 33 } 34 35 bool cmpy(const Node& lhs, const Node& rhs) { 36 return lhs.y < rhs.y; 37 } 38 39 int L, H, n; 40 int calc() { 41 int ans = 0; 42 for(int i = 0; i < n; i++) { 43 int U = H, D = 0; 44 for(int j = i + 1; j < n; j++) { 45 ans = max(ans, (p[j].x - p[i].x) * (U - D)); 46 if(D <= p[j].y && p[j].y <= U) { 47 if(p[j].y > p[i].y) U = p[j].y; 48 else D = p[j].y; 49 } 50 if(U == D) break; 51 } 52 } 53 return ans; 54 } 55 56 int main() { 57 #ifdef DEBUG 58 freopen("in.txt", "r", stdin); 59 freopen("out.txt", "w", stdout); 60 #endif 61 62 read(H), read(L), read(n); 63 for(int i = 0; i < n; i++) { 64 int x = read<int>(), y = read<int>(); 65 p[i] = Node(x, y); 66 } 67 p[n++] = Node(0, 0); 68 p[n++] = Node(0, H); 69 p[n++] = Node(L, 0); 70 p[n++] = Node(L, H); 71 72 int ans = 0; 73 sort(p, p + n, cmpy); 74 for(int i = 1; i < n; i++) { 75 ans = max(ans, (p[i].y - p[i-1].y) * L); 76 } 77 sort(p, p + n, cmpx); 78 ans = max(ans, calc()); 79 for(int i = 0; i < n; i++) p[i].x *= -1; 80 reverse(p, p + n); 81 printf("%d\n", max(ans, calc())); 82 83 return 0; 84 } View Code

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轉載于:https://www.cnblogs.com/showson/p/5097851.html

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