poj 2398 Toy Storage（计算几何：叉积）

来源：互联网发布：软件服务行业编辑：程序博客网时间：2024/05/18 02:17
基本上和poj 2318一模一样。。。
改下输出就可以了
代码如下：
/* ***********************************************Author        :yinhuaCreated Time  :2014年12月01日 星期一 19时25分15秒File Name     :poj2398.cpp************************************************ */#include <cmath>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>#define MAXN 100010#define LL long longusing namespace std;struct Point {      int x, y;      Point() { }        Point(int _x, int _y) {          x = _x;y = _y;      }        Point operator -(const Point &b) const {          return Point(x-b.x, y-b.y);      }      int operator *(const Point &b) const {          return x*b.x+y*b.y;      }      int operator ^(const Point &b) const {          return x*b.y-y*b.x;      }  };    struct Line {      Point s, e;      Line() { }      Line(Point _s, Point _e) {          s = _s; e = _e;      }  };  bool cmp(Line a, Line b) {    return (a.s).x < (b.s).x;}  int xmult(Point p0, Point p1, Point p2) {//p0点与线段p1 p2的叉积      return (p1-p0)^(p2-p0);  }    Line line[MAXN];  int res[MAXN];int ans[MAXN];    int main(void) {      int m, n, x1, y1, x2, y2;      bool first = true;      while(~scanf("%d", &n) && n) {            scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);          int ui, li;          for(int i=0; i<n; ++i) {              scanf("%d%d", &ui, &li);              line[i] = Line(Point(ui, y1), Point(li, y2));          }          sort(line, line+n, cmp);        line[n] = Line(Point(x2, y1), Point(x2, y2));          int x, y;          Point p;          memset(ans, 0, sizeof(ans));          for(int i=1; i<=m; ++i) {            scanf("%d%d", &x, &y);              p = Point(x, y);              int l = 0, r = n;              int tmp;              while(l <= r) {                  int mid = (l+r)>>1;                  if(xmult(p, line[mid].s, line[mid].e) < 0) {//点在当前线的左侧                      tmp = mid;                      r = mid-1;                  } else l = mid+1;//点在当前线的右侧              }              ans[tmp]++;          }          memset(res, 0, sizeof(res));        for(int i=0; i<=n; ++i) {            if(ans[i]) ++res[ans[i]];        }        bool ok = true;        for(int i=1; i<=m; ++i) {            if(res[i]) {                if(ok) {                    ok = false;                    puts("Box");                }                printf("%d: %d\n", i, res[i]);            }        }    }      return 0;  }
0 0