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题目大意：有n棵树，每棵树有坐标(x,y)，价值v，长度l，问如何砍能砍掉最小价值为的树（价值相同则砍最少的树），能把其他树都围起来
思
路：枚举所有砍树的方案（我用的递归，用二进制的方法理论上来说也可以），算一下能不能围起剩下的树（如果价值比当前答案要大就不用算了）。至于怎么围起
剩下的树，一个点的明显是需要0长度，两个点就需要这两个点的距离*2，三个点或以上就要用到求凸包的方法（反正我的凸包是不能算三个点以下的）
PS：输出最好复制啊，我好像就是因为forest打错了WA了好几次啊……
AC Code:
1 #include <cstdio> 2 #include <cmath> 3 #include <algorithm> 4 using namespace std; 5 6 const double EPS = 1e-6; 7 8 inline int sgn(const double &x) { 9 if(fabs(x) < EPS) return 0; 10 return x > 0 ? 1 : -1; 11 } 12 13 struct Point { 14 double x, y; 15 int v, l; 16 }; 17 18 inline bool Cross(Point &sp, Point &ep, Point &op) { 19 return (sp.x - op.x) * (ep.y - op.y) - (ep.x - op.x) * (sp.y - op.y) >= 0; 20 } 21 22 inline double dist(Point &a, Point &b) { 23 return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); 24 } 25 26 inline bool cmp(const Point &a, const Point &b) { 27 if(a.y == b.y) return a.x < b.x; 28 return a.y < b.y; 29 } 30 31 const int MAXN = 20; 32 int stk[MAXN]; 33 bool cut[MAXN], ans[MAXN]; 34 Point p[MAXN], a[MAXN]; 35 int n, top; 36 double answood; 37 38 double Graham_scan(int n) { 39 sort(p, p + n, cmp); 40 top = 1; 41 stk[0] = 0; stk[1] = 1; 42 for(int i = 2; i < n; ++i) { 43 while(top && Cross(p[i], p[stk[top]], p[stk[top - 1]])) --top; 44 stk[++top] = i; 45 } 46 int len = top; 47 stk[++top] = n - 2; 48 for(int i = n - 3; i >= 0; --i) { 49 while(top != len && Cross(p[i], p[stk[top]], p[stk[top - 1]])) --top; 50 stk[++top] = i; 51 } 52 double sum = 0; 53 stk[++top] = stk[0]; 54 for(int i = 0; i < top; ++i) 55 sum += dist(p[stk[i]], p[stk[i+1]]); 56 return sum; 57 } 58 59 int minval, mincut, sumval, sumlen; 60 double uselen; 61 62 void setans(int cutcnt) { 63 for(int i = 1; i <= n; ++i) ans[i] = cut[i]; 64 minval = sumval; 65 mincut = cutcnt; 66 answood = sumlen - uselen; 67 } 68 69 void dfs(int dep, int cutcnt) { 70 if(dep == n + 1) { 71 if(n == cutcnt) return ; 72 sumval = sumlen = 0; 73 for(int i = 1; i <= n; ++i) { 74 if(!cut[i]) continue; 75 sumval += a[i].v; 76 sumlen += a[i].l; 77 } 78 if(sumval > minval) return ; 79 if(sumval == minval && cutcnt >= mincut) return ; 80 if(n - cutcnt == 1) { 81 uselen = 0; 82 setans(cutcnt); 83 } 84 else if(n - cutcnt == 2) { 85 int i1 = 0, i2 = 0; 86 for(int i = 1; i <= n; ++i) { 87 if(cut[i]) continue; 88 if(!i1) i1 = i; 89 else i2 = i; 90 } 91 uselen = 2 * dist(a[i1], a[i2]); 92 if(uselen <= sumlen) setans(cutcnt); 93 } 94 else { 95 int pcnt = 0; 96 for(int i = 1; i <= n; ++i) { 97 if(cut[i]) continue; 98 p[pcnt++] = a[i]; 99 } 100 uselen = Graham_scan(pcnt); 101 if(sgn(uselen - sumlen) <= 0) setans(cutcnt); 102 } 103 return ; 104 } 105 cut[dep] = false; 106 dfs(dep + 1, cutcnt); 107 cut[dep] = true; 108 dfs(dep + 1, cutcnt + 1); 109 } 110 111 int main() { 112 int ca = 1; 113 while(scanf("%d", &n) != EOF && n) { 114 for(int i = 1; i <= n; ++i) { 115 scanf("%lf%lf%d%d", &a[i].x, &a[i].y, &a[i].v, &a[i].l); 116 } 117 mincut = MAXN; 118 minval = 0x7fffffff; 119 dfs(1, 0); 120 if(ca != 1) printf(" "); 121 printf("Forest %d ", ca++); 122 printf("Cut these trees:"); 123 for(int i = 1; i <= n; ++i) if(ans[i]) printf(" %d", i); 124 printf(" Extra wood: %.2f ", answood); 125 } 126 }
By 区彦开