题意：给定一些树木的坐标及砍掉树木所能围成的长度，还有他的价值，问在损失最少的价值的情况下砍掉那些树，使得砍掉的树足够围住剩下的树。。
思路：枚举砍掉的树，然后进行一次凸包，判断即可
code：
1 /* 2 State: Accepted 3 Time: 2013-03-30 17:27:14 4 */ 5 6 #include <iostream> 7 #include <cmath> 8 #include <cstring> 9 #include <cstdlib> 10 #include <cmath> 11 #include <algorithm> 12 #include <cstdio> 13 #include <set> 14 #define M0 memset(a, 0 ,sizeof(a)) 15 struct oo {int x ,y , l, v;}; 16 using namespace std; 17 oo a[100], b[100], q[100]; 18 int n , tot, ansv , ansopt , px , py , test = 0,ansnum; 19 double anslen; 20 void init(){ 21 anslen = 0; 22 ansv = ansopt = 0; 23 ansnum = 0x7fffffff; 24 memset(a, 0 ,sizeof(a)); 25 for (int i = 1; i <= n; ++i) 26 scanf("%d%d%d%d",&a[i].x, &a[i].y, &a[i].v ,&a[i].l); 27 28 } 29 30 bool cmp(const oo a, const oo b){ 31 if (a.x < b.x|| a.x == b.y && a.y < b.y ) return true; 32 return false; 33 } 34 35 bool cmpx(const oo a, const oo b){ 36 int x1 = a.x - px; 37 int x2 = b.x - px; 38 int y1 = a.y - py; 39 int y2 = b.y - py; 40 if (x1 * y2 -x2* y1 > 0) return true; 41 if (x1*y2 - x2*y1 == 0 && x1*x1 + y1*y1 < x2*x2 +y2*y2) return true; 42 return false; 43 } 44 45 void solve(int opt){ 46 tot = 0; 47 oo c; 48 memset(b, 0, sizeof(b)); 49 int value = 0 , len = 0; 50 int x1 , y1, x2, y2; 51 double x,y; 52 for (int i = 1; i <= n; ++i){ 53 if (((1 << i) & opt )!= 0){ 54 b[++tot] = a[i]; 55 value += a[i].v; 56 } 57 else len += a[i].l; 58 } 59 if (value < ansv) return; 60 sort(b+1, b+1+tot, cmp); 61 for (int i = 2; i < tot; ++i) 62 for (int j = i + 1; j <= tot; ++j ) 63 if (!cmpx(a[i], a[j])) { 64 c = a[i]; 65 a[i] = a[j]; 66 a[j] = c; 67 } 68 // if (tot > 2) sort(b + 2, b + tot + 1, cmpx); 69 memset(q, 0 ,sizeof(q)); 70 q[1] = b[1]; 71 q[2] = b[2]; 72 int h = 1 , t = 2; 73 for (int i = 3; i <= tot; ++i){ 74 while (t > h){ 75 x1 = b[i].x - q[t-1].x; 76 x2 = q[t].x - q[t-1].x; 77 y1 = b[i].y - q[t-1].y; 78 y2 = q[t].y - q[t-1].y; 79 if (x1*y2-x2*y1 > 0) -- t; 80 else break; 81 } 82 q[++t] = b[i]; 83 } 84 if (tot == 1) t = 1; 85 q[++t] = b[1]; 86 double len1 = 0; 87 for (int i = 2; i <= t; ++i){ 88 x = q[i].x - q[i-1].x; 89 y = q[i].y - q[i-1].y; 90 len1 += (pow(x*x+y*y,0.5)); 91 } 92 93 if (len >= len1){ 94 if (ansv < value || ansv == value && ansnum > n - tot){ 95 ansv = value; 96 ansopt = opt; 97 anslen = len - len1; 98 ansnum = n - tot; 99 } 100 } 101 } 102 103 void dfs(int i, int opt){ 104 if (i > n) { 105 solve(opt); 106 return; 107 } 108 dfs(i + 1, opt); 109 dfs(i + 1, opt |(1 << i) ); 110 111 } 112 113 void print(){ 114 printf("Forest %d\nCut these trees: ",test); 115 for (int i =1; i <= n; ++i) 116 if (((1 << i)&ansopt) == 0) 117 printf("%d ",i); 118 printf("\nExtra wood: %.2lf\n\n",anslen); 119 120 } 121 122 int main(){ 123 freopen("poj1873.in","r",stdin); 124 freopen("poj1873.out","w",stdout); 125 while ( scanf("%d",&n) != EOF && n != 0) 126 { 127 ++test; 128 init(); 129 dfs(1, 0); 130 print(); 131 } 132 fclose(stdin); fclose(stdout); 133 }