hdu4923 Room and Moor 单调栈

Room and Moor

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 276 Accepted Submission(s): 79



[align=left]Problem Description[/align]
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:



[align=left]Input[/align]
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.

[align=left]Output[/align]
Output the minimal f (A, B) when B is optimal and round it to 6 decimals.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

1.428571
1.000000
0.000000
0.000000

多校第6场，前两题出的还挺快，后面就呵呵了=。=

先把前面的1和后面的0去掉。

我当时想的是会不会就取0...x...1这样子，结果不对。。

正确答案是0...x1...x2...x3....1，中间可能有很多取值，满足x1<x2<x3...先把剩下的序列分成一段一段的，连续的1和连续的0为一段，求一个x，为什么连续的1和连续的0为一段呢，因为如果把连续1看成一段，连续0看成一段，1的取值是1，x的取值是0，0<1，所以肯定还是要合并的。为什么这一段只取一个x呢，因为1的段希望x尽量大，0的段希望x尽量小，0段的取值又不能小于1段的取值，所以相等是最优了。x的算法根据抛物线，-b/(2a)。每段求出来的x放到栈里，如果比上段的x小，就要和上段合并，把这段和上段看成一段，再用同样的方法算x，如果又比上段小，继续合并。。

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<set>
#include<map>
#include<stack>
#define INF 0x3f3f3f3f
#define MAXN 100010
#define MAXM 1500
#define MOD 1000000007
#define MAXNODE 8*MAXN
#define eps 1e-10
using namespace std;
typedef long long LL;
int T,N,a[MAXN],cnt1[MAXN];
struct seg{
int s,e;
double x;
};
stack<seg> st;
int main(){
freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i=1;i<=N;i++) scanf("%d",&a[i]);
int p1,p2,flag1=0,flag2=0;
for(int i=1;i<=N;i++) if(a[i]!=0){
p1=i;
flag1=1;
break;
}
if(!flag1){
printf("0.000000\n");
continue;
}
for(int i=N;i>=1;i--) if(a[i]!=1){
p2=i;
flag2=1;
break;
}
if(!flag2){
printf("0.000000\n");
continue;
}
if(p1==p2+1){
printf("0.000000\n");
continue;
}
cnt1[0]=0;
for(int i=1;i<=N;i++){
cnt1[i]=cnt1[i-1];
if(a[i]==1) cnt1[i]=cnt1[i-1]+1;
}
while(!st.empty()) st.pop();
int s,e,i=p1;
while(i<=p2){
for(s=i;s<=p2&&a[s]==1;s++);
for(e=s;e+1<=p2&&a[e+1]==0;e++);
double A=e-i+1,B=-2*(cnt1[e]-cnt1[i-1]),x=(-B)/(2*A);
int ls=i;
while(!st.empty()&&x<st.top().x){
ls=st.top().s;
st.pop();
A=e-ls+1,B=-2*(cnt1[e]-cnt1[ls-1]),x=(-B)/(2*A);
}
st.push((seg){ls,e,x});
i=e+1;
}
double ans=0;
while(!st.empty()){
seg tmp=st.top();
st.pop();
double A=tmp.e-tmp.s+1,B=-2*(cnt1[tmp.e]-cnt1[tmp.s-1]),C=cnt1[tmp.e]-cnt1[tmp.s-1];
ans+=(4*A*C-B*B)/(4*A);
}
printf("%.6lf\n",ans);
}
return 0;
}