poj 2398 Toy Storage

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题意：跟2318一样的，就是给你的纸板没有排序而已。输出为包含这个玩具数的区域有多少个，按升序输出。例如测试数据2: 5，表示最终结果有5个区域的玩具数为2。

分析：将输入的纸板排序后，然后输出改一下即可。

以下附上代码：

#include <algorithm>#include <iostream>#include <cstdio>using std::cout;using std::sort;const int maxn = 1005;struct Point{  int x,y;    Point(){}    Point(int _x, int _y) :    x(_x), y(_y) {}};struct Vec{  Point a,b;    Vec(){}    Vec(Point _a, Point _b) :    a(_a), b(_b) {}    Vec(const Vec &t){    a = t.a; b = t.b;  }    bool operator < (const Vec &t) const{    return a.x < t.a.x;  }};int n,m;int x1,y1,x2,y2;Vec box[maxn]; //纸板 int cnt[maxn]; //toys cnt int res[maxn]; //结果 int Cross(Vec a, Point t){  Vec b(a.a,t);  int ax = a.b.x - a.a.x;  int ay = a.b.y - a.a.y;  int bx = b.b.x - b.a.x;  int by = b.b.y - b.a.y;    return ax * by - ay * bx;}bool input(){  int tx1,tx2;  int tx,ty;  int i;  scanf("%d",&n);  if(n == 0)    return false;  scanf("%d",&m);  scanf("%d%d%d%d",&x1,&y1,&x2,&y2);  box[0] = Vec(Point(x1, y1), Point(x1, y2));  box[n+1] = Vec(Point(x2, y2), Point(x2, y1));  for(i = 1; i <= n; i++){    scanf("%d%d",&tx1,&tx2);    box[i] = Vec(Point(tx1, y1), Point(tx2, y2));  }    sort(box,box+n+2);      for(i = 0; i <= n; i++){    res[i] = 0; cnt[i] = 0;  }    return true;}int find(Point t){  int l = 0;  int r = n+1;  int m;    while(r - l > 1){    m = l + (r - l) / 2;    if(Cross(box[m], t) < 0){      r = m;    }    else{      l = m;    }  }  return l;}void solve(){  int x,y;  int i;  for(i = 0; i < m; i++){    scanf("%d%d",&x,&y);    cnt[find(Point(x, y))]++;  }    for(i = 0; i < maxn; i++){    if(cnt[i])      res[cnt[i]]++;  }  for(i = 1; i < maxn; i++){    if(res[i]){      printf("%d: %d\n",i,res[i]);    }  }}int main(){  while(input()){    puts("Box");    solve();  }return 0;}

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