Hdu 4923（單調棧）

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題目連結

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 842    Accepted Submission(s): 250

Problem DescriptionPM 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:

 

InputThe 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.

 

OutputOutput the minimal f (A, B) when B is optimal and round it to 6 decimals.

 

Sample Input491 1 1 1 1 0 0 1 191 1 0 0 1 1 1 1 140 0 1 140 1 1 1

 

Sample Output1.4285711.0000000.0000000.000000

 Accepted Code:

 1 /************************************************************************* 2     > File Name: 4923.cpp 3     > Author: Stomach_ache 4     > Mail: [email protected] 5     > Created Time: 2014年08月08日 星期五 22時27分38秒 6     > Propose:  7  ************************************************************************/ 8  9 #include <cmath>10 #include <string>11 #include <cstdio>12 #include <fstream>13 #include <cstring>14 #include <iostream>15 #include <algorithm>16 using namespace std;17 18 const double eps = 1e-12;19 const int maxn = 100002;20 int n;21 int a[maxn], st[maxn];22 23 struct node {24     double x; //區間均值25     int l, r, one; //區間範圍及1的個數26 }A[maxn];27 28 // unite i to j29 void unite(int i, int j) {30       A[j].l = A[i].l;31     A[j].x = A[j].one + A[i].one + 0.0;32     A[j].x /= A[j].r - A[j].l + 1;33     A[j].one = A[i].one + A[j].one;34 }35 36 double solve() {37       int len = 1;38     for (int i = 1; i <= n; i++) {39           A[len].x = a[i] + 0.0;40         A[len].l = i;41         i++;42         while (i <= n && a[i] == a[i-1]) i++;43         A[len].r = i - 1;44         i--;45         if (a[i]) A[len].one = A[len].r - A[len].l + 1;46         else A[len].one = 0;47         len++;48     } 49     int top = 0;50     for (int i = 1; i < len; i++) {51           while (top && A[i].x < A[st[top-1]].x) {52               unite(st[top-1], i);53             top--;54         }55         st[top++] = i;56     }57     double ans = 0.0;58     for (int i = 0; i < top; i++) {59           for (int j = A[st[i]].l; j <= A[st[i]].r; j++) {60               ans += (a[j] - A[st[i]].x) * (a[j] - A[st[i]].x);61         }62     }63     return ans + eps;64 }65 66 int main(void) {67       int t;68     scanf("%d", &t);69     while (t--) {70           scanf("%d", &n);71         for (int i = 1; i <= n; i++) scanf("%d", a + i);72         printf("%.6f\n", solve());73     }74     return 0;75 }