POJ 1873（计算几何初步——位运算枚举，求凸包）

这道题很水很水，由于数据范围很小，所以根本不用搜索。

直接位运算枚举就行了。不过要注意一点：

凸包点数小于3的情况需特判。

好了，贴代码....

View Code

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 #include<vector>
  6 #include<map>
  7 #include<set>
  8 #include<cmath>
  9 #include<complex>
 10 #include<algorithm>
 11 #define EPS 1e-10
 12 #define largty(x,y) (x-EPS)>(y)
 13 #define smalty(x,y) (x+EPS)<(y)
 14 #define nlargty(x,y) (x-EPS)<(y)
 15 #define nsmalty(x,y) (x+EPS)>(y)
 16 #define equal(x,y) (nlargty(x,y) && (nsmalt(x,y)))
 17 #define max(a,b) (a)>(b)?(a):(b)
 18 #define min(a,b) (a)<(b)?(a):(b)
 19 #define MAXN 20 
 20 using namespace std;
 21 struct point {
 22     double x,y;
 23     int value;
 24     double length;
 25 }tree[MAXN],convex[MAXN],data[MAXN];
 26 point start;
 27 int T;
 28 
 29 inline double p_distance(point &p, point &q)
 30 {
 31     return sqrt(pow(q.x - p.x,2) + pow(q.y - p.y,2));    
 32 }
 33 
 34 inline  double cross(const point &p1, const point &p2, const point &q1, const point &q2)
 35 {
 36     return (q2.y - q1.y)*(p2.x - p1.x) - (q2.x - q1.x)*(p2.y - p1.y);    
 37 }
 38 
 39 inline bool cmp1(const point &p, const point &q)//选择Y-X排序下最小的点 
 40 {
 41     return p.y < q.y || (p.y == q.y && p.x < q.x);    
 42 }
 43 
 44 inline bool cmp2(const point &a, const point &b)//逆时针排序，极角相同则距离短的排前面 
 45 {
 46     point origin;
 47     origin = start;
 48     return cross(origin,a,origin,b) > 0 || (cross(origin,a,origin,b) == 0 && fabs(a.x) < fabs(b.x));
 49 }
 50 
 51 inline void convex_hull(int &cnt, int n)
 52 {
 53     sort(data, data+n, cmp1);
 54     start = data[0];
 55     sort(data+1, data+n, cmp2);
 56     convex[cnt++] = start;
 57     convex[cnt++] = data[1];
 58     for (int j(2); j<n; ++j) {
 59         while (cnt >= 2 && cross(convex[cnt-2],convex[cnt-1],convex[cnt-1],data[j]) <= 0)--cnt;//共线的话也应该弹出原来栈中的点 
 60             convex[cnt++] = data[j];    
 61     }    
 62 }
 63 
 64 inline double get_circum(int cnt)
 65 {
 66     double circum = 0;
 67     for (int i(0); i<cnt; ++i) {
 68         circum += p_distance(convex[i],convex[(i+1)%cnt]);
 69     }
 70     return circum;    
 71 }
 72 
 73 inline void get_tree(int data)
 74 {
 75     printf("Cut these trees:");
 76     for (int i(0); data != 0; ++i,data >>= 1) {
 77         if (data%2 == 1) {
 78             printf(" %d",i+1);    
 79         }
 80     }        
 81 }
 82 
 83 inline int enum_tree(int n, double &end_circum)
 84 {
 85     double circum = 0;
 86     int result = 0;
 87     int bound = (1<<n) - 1;
 88     int min_value = INT_MAX;
 89     int min_tree = INT_MAX;
 90     for (int i(1); i<bound; ++i) {
 91         int j = 0;
 92         int temp = i;
 93         int tree_num = 0;
 94         int value = 0;
 95         double length = 0;
 96         for (int cnt(0); cnt<T; ++cnt,temp >>= 1) {
 97             if (temp != 0 && temp%2 == 1) {
 98                 value += tree[cnt].value;
 99                 length += tree[cnt].length;                
100                 ++tree_num;
101             } else data[j++] = tree[cnt];    
102         }
103         if (value > min_value)continue;
104         if (j >= 2) {
105         int top = 0;
106             convex_hull(top,j);
107             circum = get_circum(top);
108         } else {
109             circum = 0;    
110         }
111         if (length >= circum) {
112             if (value == min_value) {
113                 if (tree_num < min_tree) {
114                     min_tree= tree_num;
115                     result = i;    
116                     end_circum = circum;
117                 }
118             } else {
119                 if (value < min_value) {
120                     min_value = value;
121                     min_tree = tree_num;
122                     result = i;
123                     end_circum = circum;
124                 }    
125             }
126         }
127     }
128     return result;    
129 }
130 
131 inline double length(int n)
132 {
133     double len = 0;
134     for (int i(0); n != 0; ++i,n >>= 1) {
135         if (n%2 == 1) {
136             len += tree[i].length;    
137         }    
138     }
139     return len;
140 }
141 
142 int main()
143 {
144     int count = 1;
145     while (scanf("%d",&T) && T != 0) {
146         for (int i(0); i<T; ++i) {
147             scanf("%lf%lf%d%lf",&tree[i].x,&tree[i].y,&tree[i].value,&tree[i].length);
148         }
149         double circum = 0;
150         int result = enum_tree(T,circum);
151         printf("Forest %d\n",count++); 
152         get_tree(result);
153         printf("\nExtra wood: %.2lf\n\n",length(result) - circum);    
154     }
155     return 0;    
156 }