Room and Moor
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 557 Accepted Submission(s):
159
Problem Description
PM Room defines a sequence A = {A1,
A2,..., AN}, each of which is either 0 or 1. In order to
beat him, programmer Moor has to construct another sequence B = {B1,
B2,... , BN} of the same length, which satisfies
that:
Input
The input consists of multiple test cases. The number
of test cases T(T<=100) occurs in the first line of input.
For each test case:
The first line contains a single integer N (1<=N<=100000), which denotes the length of A and B.
The second line consists of N integers, where the ith denotes Ai.
For each test case:
The first line contains a single integer N (1<=N<=100000), which denotes the length of A and B.
The second line consists of N integers, where the ith denotes Ai.
Output
Output the minimal f (A, B) when B is optimal and round
it to 6 decimals.
Sample Input
4
9
1 1 1 1 1 0 0 1 1
9
1 1 0 0 1 1 1 1 1
4
0 0 1 1
4
0 1 1 1
Sample Output
1.428571
1.000000
0.000000
0.000000
思路：首先去掉前导零和最后的1，相当于把整个序列分成几个区间，每部分以1开头，0结尾，即如1 0 1 1 0 0等，可知对于每一个区间，要取得最小值，那这个部分所有的值即对应的这个区间内的平均数，如果这个平均数和前面一个区间的相比较大，就压入栈，否则将栈里的元素顶出，并与当前区间合并求平均数……知道比前面的大为止，最后求出每个区间的对应的Seg(ai - bi)^2 就可以了。
1 #include <cstdio> 2 #include <cstring> 3 #include <stack> 4 #define eps 0.00000001 5 using namespace std; 6 7 const int LEN = 100010; 8 int arr[LEN]; 9 struct line 10 { 11 int l, r, sum; 12 double rate; 13 }; 14 stack<line> s; 15 int main() 16 { 17 int T, n; 18 line tmp; 19 scanf("%d", &T); 20 while(T--) 21 { 22 scanf("%d", &n); 23 for(int i = 0; i < n; i++) 24 scanf("%d", &arr[i]); 25 int h = 0; 26 while(arr[h] == 0) 27 h++; 28 int k = n - 1; 29 while(arr[k] == 1) 30 k--; 31 for(int i = h; i <= k; i++) 32 { 33 if (i == h || i > h && arr[i-1] == 0 && arr[i] == 1) 34 { 35 tmp.l = i; 36 tmp.sum = 0; 37 } 38 if (i < k && arr[i] == 0 && arr[i+1] == 1 || i == k) 39 { 40 tmp.r = i; 41 tmp.rate = tmp.sum * 1.0 / ((tmp.r - tmp.l + 1) * 1.0); 42 while(true) 43 { 44 if (s.empty() || s.top().rate - tmp.rate < eps) 45 { 46 s.push(tmp); 47 break; 48 } 49 if (s.top().rate - tmp.rate > eps) 50 { 51 tmp.l = s.top().l; 52 tmp.sum += s.top().sum; 53 tmp.rate = tmp.sum*1.0 / ((tmp.r - tmp.l + 1)*1.0); 54 s.pop(); 55 } 56 } 57 } 58 if (arr[i] == 1) 59 tmp.sum++; 60 } 61 double ans = 0; 62 while(!s.empty()) 63 { 64 ans += ((1 - s.top().rate) * (1 - s.top().rate) * s.top().sum + s.top().rate * s.top().rate * (s.top().r - s.top().l + 1 - s.top().sum)); 65 s.pop(); 66 } 67 printf("%f ", ans); 68 } 69 return 0; 70 }