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 60000
pair<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;
			else
				up = 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;
			else
				up = 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;
}