Greedy:Bound Found(POJ 2566)
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题目大意:就是给你一个数组,要你找出连续的数的绝对值的和最接近t的那一串,并且要找出数组的上界和下界的下标,并显示他们的和
因为这一题的数有正有负,所以必须要先把和求出来,然后排序,然后利用a(s,t)=sum(t)-sum(s)找出目标
1 #include <iostream> 2 #include <algorithm> 3 #include <functional> 4 5 using namespace std; 6 7 //pair<int, int>Acc[100016]; 8 static struct _set 9 { 10 int sum, index; 11 bool operator<(const _set&x)const 12 { 13 return sum < x.sum; 14 } 15 }Acc[100002]; 16 17 void solve(const int, const int); 18 int get_sum(int *const, int*const, int*const, const int, const int,const int); 19 int ABS(int); 20 21 int main(void)//游标卡尺大法 22 { 23 int n, k, t, tmp; 24 25 while (~scanf("%d%d", &n, &k)) 26 { 27 if (n == 0 && k == 0) 28 break; 29 Acc[0].sum = 0; Acc[0].index = 0; 30 for (int i = 1; i <= n; i++) 31 { 32 scanf("%d", &tmp); 33 Acc[i].index = i; Acc[i].sum = Acc[i - 1].sum + tmp; 34 } 35 sort(Acc, Acc + n + 1);//直接给和排序 36 37 for (int i = 0; i < k; i++) 38 { 39 scanf("%d", &t); 40 solve(n, t); 41 } 42 } 43 return 0; 44 } 45 46 void solve(const int n, const int S) 47 { 48 int ans_sum, ans_lb, ans_ub, lb, ub, sum; 49 50 lb = ub = 0; sum = ans_sum = 0x80808080; 51 while (1) 52 { 53 while (ub < n && sum < S)//标准尺取法 54 sum = get_sum(&ans_sum, &ans_lb, &ans_ub, lb, ++ub, S); 55 if (sum < S) 56 break; 57 sum = get_sum(&ans_sum, &ans_lb, &ans_ub, ++lb, ub, S); 58 } 59 printf("%d %d %d\n", ans_sum, ans_lb + 1, ans_ub); 60 } 61 62 int ABS(int x) 63 { 64 return x >= 0 ? x : -x; 65 } 66 67 int get_sum(int *const ans_sum, int*const ans_lb, int*const ans_ub, const int lb, const int ub,const int S) 68 { 69 if (lb >= ub) 70 return INT_MIN; 71 int tmp = Acc[ub].sum - Acc[lb].sum; 72 if (ABS(tmp - S) < ABS(*ans_sum - S)) 73 { 74 *ans_sum = tmp; 75 *ans_lb = min(Acc[ub].index, Acc[lb].index); 76 *ans_ub = max(Acc[ub].index, Acc[lb].index); 77 } 78 return tmp; 79 }
参考http://www.hankcs.com/program/algorithm/poj-2566-bound-found.html
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