poj1228

题目大意：给定一些点，问存不存在稳定凸包。。所谓的稳定凸包，就是每条凸包边上至少3个点。。

思路:凸包，然后判断是否有其他点在边上。。一般来说凸包算法是不含共线情况的，况且边不多，所以只能这样判断了。。

code：

  

  1 /*
  2    Time:2013-04-06 11:55:37
  3    State:Accepted
  4 
  5 */
  6 
  7 #include<iostream>
  8 #include<fstream>
  9 #include<cstring>
 10 #include<cstdlib>
 11 #include<cstdio>
 12 #include<string>
 13 #include<cmath>
 14 #include<algorithm>
 15 struct oo{int x , y; };
 16 using namespace std;
 17 int T, n, px, py ,bo[2000];
 18 oo a[2000],q[2000];
 19 
 20 void init(){
 21      scanf("%d",&n);
 22      memset(a , 0 ,sizeof(a));
 23      memset(q , 0 ,sizeof(q));
 24      memset(bo ,0 ,sizeof(bo));
 25      for (int  i = 1; i <= n ;++i) 
 26         scanf("%d%d",&a[i].x, &a[i].y); 
 27      
 28 }
 29 
 30 bool cmp(const oo a , const oo b){
 31     if (a.y < b.y) return true;
 32     if (a.y == b.y && a.x < b.x) return true;
 33     return false;
 34      
 35 }
 36 
 37 bool cmpx(const oo a, const oo b){
 38      int x1 = a.x - px;
 39      int x2 = b.x - px;
 40      int y1 = a.y - py;
 41      int y2 = b.y - py;
 42      if (x1*y2-x2*y1 > 0) return true;
 43      if (x1*y2 == x2*y1 && x1*x1 + y1*y1 < x2*x2 + y2*y2) return true;
 44      return false;
 45 }
 46 void solve(){
 47      if  (n < 6){
 48           printf("NO\n");
 49           return;    
 50      }
 51      sort(a + 1, a + 1 + n, cmp);
 52      px = a[1].x; 
 53      py = a[1].y; 
 54      if (n > 2) sort(a + 2, a + 1 + n, cmpx);
 55      int h = 1, t = 2;
 56      q[1] = a[1];
 57      q[2] = a[2];
 58      int x1, y1, x2, y2;
 59      for (int i = 3; i <= n; ++i){
 60           while (true){
 61                if (t == 1) break;
 62                x1 = a[i].x - q[t - 1].x;
 63                x2 = q[t].x - q[t - 1].x; 
 64                y1 = a[i].y - q[t - 1].y;
 65                y2 = q[t].y - q[t - 1].y;
 66                if (x1*y2 - x2*y1 >= 0) --t;
 67                else break;
 68           }
 69           q[++t] = a[i];
 70      }
 71      
 72      if (t < 3){
 73          printf("NO\n");
 74          return;
 75      }
 76      
 77      q[t + 1] = q[1];
 78      int cnt = 0;
 79      
 80      for (int i = 1; i <= t; ++i){
 81          for (int j = 1; j <= n ; ++j)
 82            if ( (a[j].x!=q[i].x || a[j].y != q[i].y) 
 83                && (a[j].x != q[i + 1].x || a[j].y != q[i + 1].y)){
 84                   if ((q[i].x - a[j].x)*(q[i + 1].y - a[j].y) != 
 85                       (q[i + 1].x - a[j].x)*(q[i].y - a[j].y)) continue;
 86                   ++cnt;
 87                   break;
 88               }
 89      }
 90      if (cnt >= t) printf("YES\n");
 91      else printf("NO\n");
 92          
 93 }
 94 
 95 int main(){
 96      freopen("poj1228.in","r",stdin);
 97      freopen("poj1228.out","w",stdout); 
 98      scanf("%d",&T);
 99      for (int i = 1;  i <= T; ++i){
100           init();
101           solve();    
102      }   
103      fclose(stdin); fclose(stdout);
104 }

 

 

posted on 2013-04-11 23:28  yzcstc  阅读(205)  评论(0)  编辑  收藏  举报