hdu 4923 Room and Moor

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 446    Accepted Submission(s): 125

Problem Description

PM Room defines a sequence A = {A₁, A₂,..., A_N}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B₁, B₂,... , B_N} 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 A_i.

 

Output

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

#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<list>#include<map>#include<set>using namespace std;const double eps=1e-9;int n;int a[100010];struct node{    int cnt;    int all;    double b;    node(int cnt=0,int all=0,double b=0):cnt(cnt),all(all),b(b){}};vector<node>cur;node res[100010];void deal_pre(){    cur.clear();    while(n>0 && a[n]==1) n--;    for(int i = 1; i <= n; i++){        if(a[i] == 1){            int j = i;            int cnt = 0;            while(j<=n && a[j]==1) j++,cnt++;            while(j<=n && a[j]==0) j++;            double b = cnt * 1.0 / (j-i);            cur.push_back(node(cnt, j-i, b));            i=j-1;        }    }}int sign(double x){    if(x > eps) return 1;    else if(x < -eps) return -1;    return 0;}//int binary(int ed, double x){//    int st = 0, mid ;//    while(st + 1 < ed){//        mid = ( st + ed ) >> 1;//        if( sign( x - res[mid].b )>= 0 ) st = mid;//        else ed = mid;//    }//    if( sign( x - res[st].b) == 0 ) st++;//    return st;//}int deal_res(int len){    int i = 0;    res[0] = cur[i++];    int ed = 0;    while(i < len){        if(sign( cur[i].b - res[ed].b ) >= 0){            res[++ed] = cur[i];        }        else{//            int st = binary(ed,cur[i].b);//            for(int iy = st; iy <= ed; iy++){//                sum += res[iy].all;//                cnt += res[iy].cnt;//            }//            ed = st;            int sum = cur[i].all;            int cnt = cur[i].cnt;            while(ed >= 0){                sum += res[ed].all;                cnt += res[ed].cnt;                double tp = 1.0 * cnt / sum;                if( ed == 0){                    res[ed]= node(cnt,sum,tp);                    break;                }                else{                    if(tp >= res[ed-1].b){                        res[ed] = node(cnt , sum, tp);                        break;                    }                    else ed--;                }            }        }        i++;    }    return ed;}double work(){    int len = cur.size();    if(len == 0) return 0;    int ed = deal_res(len);    double ans = 0;    for(int i = 0; i <= ed; i++){        int all = res[i].all;        int cnt = res[i].cnt;        double b1 = (all - cnt) * 1.0 / all;        double b2 = cnt * 1.0 / all;        ans = ans + cnt * b1 * b1 + (all - cnt) * b2 * b2;    }    return ans;}int main(){//freopen("in.in","r",stdin);    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for(int i = 1; i <= n; i++)            scanf("%d",&a[i]);        deal_pre();        printf("%.6f\n",work());    }return 0;}