简单几何(线段覆盖) POJ 3347 Kadj Squares

 

题目传送门

题意：告诉每个矩形的边长，它们是紧贴着的，问从上往下看，有几个还能看到。

分析：用网上猥琐的方法，将边长看成左端点到中心的距离，这样可以避免精度问题。然后先求出每个矩形的左右端点，然后如果被覆盖那么将端点更新到被覆盖的位置。最后看那些更新后左端点小于右端点，这些是可以看得到的。

 

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/************************************************ * Author        :Running_Time * Created Time  :2015/10/28 星期三 11:48:32 * File Name     :POJ_3347.cpp  ************************************************/   #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std;   #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = acos (-1.0); struct Square    {     int l, r, len; }s[55];   int main(void)    {     int n;     while (scanf ("%d", &n) == 1)   {         if (!n) break;         for (int i=1; i<=n; ++i)    {             scanf ("%d", &s[i].len);             s[i].l = 0;             for (int j=1; j<i; ++j) {                 int tmp;                 if (s[i].len <= s[j].len)    {                     tmp = s[j].l + s[j].len + s[i].len;                 }                 else    {                     tmp = s[j].l + s[j].len * 3 - s[i].len;                 }                 if (tmp > s[i].l)   s[i].l = tmp;             }             s[i].r = s[i].l + s[i].len * 2;         }         for (int i=2; i<=n; ++i)    {             for (int j=1; j<i; ++j) {                 if (s[j].len < s[i].len && s[j].r > s[i].l) {                     s[j].r = s[i].l;                 }                 else if (s[j].len > s[i].len && s[j].r > s[i].l)    {                     s[i].l = s[j].r;                 }             }         }         for (int i=1; i<=n; ++i)    {             if (s[i].l < s[i].r)    {                 printf ("%d ", i);             }         }         puts ("");     }      //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";       return 0; }