HDU 4923 (贪心+证明)

Room and Moor

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.

 

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

题意：不多说了。

sl： 其实就是一个贪心的思想。

网上讲解很多。比赛时没搞出来。。。呵呵了。

 1 // by caonima
 2 // hehe
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <stack>
 6 #include <queue>
 7 #include <vector>
 8 #include <algorithm>
 9 using namespace std;
 const int MAX = 1e6+;
 struct node {
     int one,zero;
     double val;
 }v[MAX];
 stack<node> ans;
 int a[MAX];
 void init() {
     while(!ans.empty()) ans.pop();
 }
 int main() {
     int cas,n;
     scanf("%d",&cas);
     while(cas--) {
         scanf("%d",&n);
         for(int i=;i<=n;i++) {
             scanf("%d",&a[i]);
         }
         a[n+]=;
         int L=,R=n;
         while(a[L]==) L++;
         while(a[R]==) R--;
         if(L>=R) printf("0.000000\n");
         else {
             int id=;
             for(int i=L;i<=R;) {
                 int cnt0=,cnt1=;
                 while(a[i]==) {
                     cnt1++; i++;
                 }
                 while(a[i]==) {
                     cnt0++; i++;
                 }
                 v[id].one=cnt1; v[id].zero=cnt0;
                 v[id++].val=(double) cnt1/(double)(cnt0+cnt1);
             }
             init();
             for(int i=;i<id;i++) {
 
                 if(ans.empty()) ans.push(v[i]);
                 else {
                     node vn=ans.top();
                     if(vn.val<=v[i].val) ans.push(v[i]);
                     else {
                         node t=v[i];
                         while(true) {
                             node vx=ans.top();
                             if(vx.val>t.val) {
                                 t.one+=vx.one; t.zero+=vx.zero;
                                 t.val=(double) t.one/(double)(t.one+t.zero);
                                 ans.pop();
                             }
                             else {
                                 ans.push(t);
                                 break;
                             }
                             if(ans.empty()) {
                                 ans.push(t);
                                 break;
                             }
                         }
                     }
                 }
             }
             double res=;
             while(!ans.empty()) {
                 node vn=ans.top();
                 ans.pop();
                 res+=vn.val*vn.val*vn.zero+(-vn.val)*(-vn.val)*vn.one;
             }
             printf("%.6lf\n",res);
         }
 
     }

84 }