Restaurant LA4851

来源：互联网发布：淘宝无线优惠券编辑：程序博客网时间：2024/05/01 22:44

一下所说的距离皆为哈密顿距离

如果一个点X对于任意的已知点Y，要么X比Y距离A近，要么X比Y距离B近，则X是一个合法点，求合法点的个数，

因为在A，B点分别有一个餐厅，而且A，B点的纵坐标相同，所以A.x左边和B.x右边的点都不合法，而假设此时只有A，B俩点，则合法点如下图所示（实点为合法点)

从中不难发现规律

现在假设一个点X在AB连线的上方，则X所限制的合法区域为

所以可以对A.x到B.x的横坐标进行扫描，找出每个横坐标出现的大于A.y最小纵坐标和小于A.y的最大横坐标，然后分别从左右俩个方向扫描合法区域，最后统计合法点数有以小细节容易出错，在注释中给出

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <queue>#include <algorithm>#include <vector>#include <cstring>#include <stack>#include <cctype>#include <utility>   #include <map>#include <string>  #include <climits> #include <set>#include <string> #include <sstream>#include <utility>#include <ctime> using std::priority_queue;using std::vector;using std::swap;using std::stack;using std::sort;using std::max;using std::min;using std::pair;using std::map;using std::string;using std::cin;using std::cout;using std::set;using std::queue;using std::string;using std::istringstream;using std::make_pair;using std::greater;const int MAXM(60010);const int INFI((INT_MAX-1) >> 1);int miy[MAXM], mxy[MAXM], Y[MAXM];int main(){int T;int m, n;scanf("%d", &T);int x1, y1, x2, y2, tx, ty;while(T--){scanf("%d%d", &m, &n);scanf("%d%d%d%d", &x1, &y1, &x2, &y2);if(x1 > x2){swap(x1, x2);swap(y1, y2);}for(int i = x1+1; i < x2; ++i){miy[i] = INFI;mxy[i] = -INFI;}for(int i = 2; i < n; ++i){scanf("%d%d", &tx, &ty);if(ty >= y1)miy[tx] = min(miy[tx], ty);if(ty <= y1)mxy[tx] = max(mxy[tx], ty);}Y[x1] = 0;for(int i = x1+1; i < x2; ++i)Y[i] = min(miy[i]-y1, y1-mxy[i]);for(int i = x1+1; i < x2; ++i)Y[i] = min(Y[i-1]+1, Y[i]);Y[x2] = 0;for(int i = x2-1; i > x1; --i)Y[i] = min(Y[i+1]+1, Y[i]);long long ans = 0LL;for(int i = x1+1; i < x2; ++i)if(Y[i]){++ans;ans += min(Y[i]-1, m-1-y1);   //注意要考率纵坐标上下限ans += min(Y[i]-1, y1);}printf("%lld\n", ans);}return 0;}