SRM 585

官网的图感觉很好，我就盗过来了。。。。（无视掉最后一张图

首先求出来每个点它向前向后最远能连到哪里l[i],r[i]

然后枚举a这个点，那么他向前最多能到达C这个点，那么对于他向后到达的b这个点，有很多情况，比如上图是三种情况

对于每种情况，我们可以看得出所形成三角形个数是r[b]-b-(C-1-b),当然r[b]就是图中的f(b)

因为b是连续的

那么对于这个式子，我们可以对r[b]-b求个sum数组，用的时候O(1)得到，后面减去的(C-1-b)这是个等差数列，也能O(1)得到。

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <ctime>#include <cstring>using namespace std;class EnclosingTriangle {public:long long getNumber(int, vector<int> , vector<int> );};#define M 60000pair<int, int> coo[20 * M];long long l[20 * M], r[20 * M];long long sumr[20 * M];struct TPoint {long long x, y;TPoint() {}TPoint(const int &_x, const int &_y) :x(_x), y(_y) {}};long long cross(const TPoint & a, const TPoint & b, const TPoint & c) {return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);}vector<int> X, Y;bool check(TPoint a, TPoint b) {for (int i = 0; i < X.size(); ++i) {if (cross(a, b, TPoint(X[i], Y[i])) < 0)return false;}return true;}long long EnclosingTriangle::getNumber(int m, vector<int> x, vector<int> y) {int n = 0;int i, j, k;int up, down, mid;X = x;Y = y;for (i = 0; i <= m; ++i) {coo[n++] = make_pair(0, i);}for (i = 1; i <= m; ++i) {coo[n++] = make_pair(i, m);}for (i = m - 1; i >= 0; --i) {coo[n++] = make_pair(m, i);}for (i = m - 1; i > 0; --i) {coo[n++] = make_pair(i, 0);}for (i = 0; i < n; ++i) {coo[i + n] = coo[i];coo[i + 2 * n] = coo[i];}for (i = n; i < 2 * n; ++i) {up = i - n + 1;down = i - 1;while (down - up > 0) {mid = (up + down) / 2;if (check(TPoint(coo[i].first, coo[i].second), TPoint(coo[mid].first, coo[mid].second)))down = mid;elseup = mid + 1;}l[i] = l[i - n] = l[i + n] = i - down;up = i + n - 1;down = i + 1;while (up - down > 0) {mid = (up + down + 1) / 2;if (check(TPoint(coo[mid].first, coo[mid].second), TPoint(coo[i].first, coo[i].second)))down = mid;elseup = mid - 1;}r[i] = r[i - n] = r[i + n] = down - i;}sumr[0] = r[0];for (i = 1; i < 3 * n; ++i)sumr[i] = r[i] + sumr[i - 1];long long ans = 0;for (i = n; i < 2 * n; ++i) {int c = i - l[i];int left = c - l[c] + n;if (left == i)left++;int right = i + r[i];if (right < left)continue;ans += sumr[right] - sumr[left - 1];ans -= (long long)(c + n - right + c + n - left - 2) * (right - left + 1) / 2;if (right == c + n)ans--;if (r[i] + r[right] == n)ans--;}return ans / 3;}