【HDU】4923 Room and Moor(2014多校第六场1003)

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 263    Accepted Submission(s): 73


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.
 

 

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
 

 

Source
2014 Multi-University Training Contest 6
 

 

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题意：很容易就读懂了。
题解：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。至于为什么。。。。说实话全是YY，居然A掉了。。
 
AC代码如下：
 
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <stack>
 4 using namespace std;
 5 
 6 #define eps 0.00000001
 7 const int LEN = 100010;
 8 
 9 int arr[LEN];
10 struct line
11 {
12     int l, r, sum;
13     double rate;
14 };
15 
16 stack<line> s;
17 
18 int main()
19 {
20     int T, n;
21     line tmp;
22     scanf("%d", &T);
23     while(T--){
24         scanf("%d", &n);
25         for(int i = 0; i < n; i++)
26             scanf("%d", arr+i);
27         int h = 0;
28         while(arr[h] == 0)
29             h++;
30         int k = n - 1;
31         while(arr[k] == 1)
32             k--;
33         for(int i = h; i <= k; i++){
34             if (i == h || i > h && arr[i-1] == 0 && arr[i] == 1){
35                 tmp.l = i;
36                 tmp.sum = 0;
37             }
38             if (i < k && arr[i] == 0 && arr[i+1] == 1 || i == k){
39                 tmp.r = i;
40                 //printf("l = %d, r = %d\n", tmp.l, tmp.r);
41                 tmp.rate = tmp.sum * 1.0 / ((tmp.r - tmp.l + 1) * 1.0);
42                 //printf("rate=%f\n", tmp.rate);
43                 while(true){
44                     if (s.empty() || s.top().rate - tmp.rate < eps){
45                         s.push(tmp);
46                         break;
47                     }
48                     if (s.top().rate - tmp.rate > eps){
49                         tmp.l = s.top().l;
50                         tmp.sum += s.top().sum;
51                         tmp.rate = tmp.sum*1.0 / ((tmp.r - tmp.l + 1)*1.0);
52                         s.pop();
53                     }
54                 }
55             }
56             if (arr[i] == 1)
57                 tmp.sum++;
58         }
59         double ans = 0;
60         while(!s.empty()){
61             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));
62             s.pop();
63         }
64         printf("%f\n", ans);
65     }
66     return 0;
67 }

 

posted @ 2014-08-07 20:29  Prime-Kv  阅读(257)  评论(0)  编辑  收藏  举报