题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5720
题意：地上有n根树枝， 长度都给出来了，你手上有长度范围为[L, R]的树枝，问有多少树枝不能和地上任意两根树枝组成三角形
思路：先sort一遍地上的树枝a[]，发现相邻两根树枝包含了两根中任意一根和剩下的树枝中任意一根的所有情况；
1 #include <algorithm> 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <string> 6 #include <cstdio> 7 #include <vector> 8 #include <ctime> 9 #include <cmath> 10 #include <queue> 11 #include <set> 12 #include <map> 13 using namespace std; 14 #define Fill(x, y) memset((x), (y), sizeof(x)) 15 #define Rep(i, x, y) for(int i = x; i <= y; ++i) 16 #define Dow(i, x, y) for(int i = x; i >= y; --i) 17 typedef long long LL; 18 typedef pair <int, int> P; 19 const int N = 1e5 + 10; 20 int GCD(int a, int b) { 21 return b ? GCD(b, a % b) : a; 22 } 23 int LCM(int a, int b) { 24 return a * b / GCD(a, b); 25 } 26 27 LL a[N]; 28 struct data { 29 LL l , r; 30 }x[N]; 31 LL cmp(data x , data y) { 32 if(x.l == y.l) 33 return x.r > y.r; 34 return x.l < y.l; 35 } 36 37 int main() 38 { 39 int t , n; 40 LL L , R; 41 scanf("%d" , &t); 42 while(t--) { 43 scanf("%d %lld %lld" , &n , &L , &R); 44 Rep(i , 1 , n) { 45 scanf("%lld" , a + i); 46 } 47 sort(a + 1 , a + n + 1); 48 x[1].l = 1, x[1].r = 0; 49 int k = 1; 50 Rep(i , 2 , n) { 51 LL x1 = a[i] - a[i - 1] + 1; 52 LL y1 = a[i] + a[i - 1] - 1; 53 if(x1 > R || y1 < L) 54 continue; 55 if(x1 < L) 56 x1 = L; 57 if(y1 > R) 58 y1 = R; 59 x[k].l = x1, x[k].r = y1; 60 k++; 61 } 62 sort(x + 1 , x + k , cmp); 63 LL res = x[1].r - x[1].l + 1; 64 // cout << res << endl; 65 LL ll = x[1].l, rr = x[1].r; 66 for(int i = 2 ; i < k ; ++i) { 67 if(x[i].l <= rr) 68 { 69 if(x[i].r > rr) 70 { 71 res += x[i].r - rr; 72 ll = x[i].l, rr = x[i].r; 73 } 74 } 75 else if(x[i].l > rr) 76 { 77 res += x[i].r - x[i].l + 1; 78 ll = x[i].l, rr = x[i].r; 79 } 80 //cout << res << endl; 81 } 82 LL res1 = R - L + 1 - res; 83 printf("%lld " , res1); 84 } 85 return 0; 86 }