hdu 4923 Room and Moor 单调栈

来源：互联网发布：怎么成为阿里云代理商编辑：程序博客网时间：2024/05/21 05:21

首先可以忽略前面的0，和后面的1.剩下的就是由 a个1 b个0 组成的01串。由不等式可知y=(a+b)*x*x-a*x+a;x=a/(a+b)时函数值最小。

而值要求递增，所以把比当前值的段都加入当前段。最后模拟输出即可。

#include <cstring>#include <algorithm>#include <iostream>#include <cstdio>#include <vector>#include <queue>#include <stack>using namespace std;#define maxn 1100000int save[maxn],n;struct node{    int len1,len0;};int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        vector<node>a;        stack<node>sta;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&save[i]);        }        save[n+1]=1;        int st=1,ed=n;        while(save[st]==0)        {            st++;        }        while(save[ed]==1)        {            ed--;        }       // printf("%d %d\n",st,ed);        if(st==ed+1)        {            printf("0.000000\n");            continue;        }        node tmp;        tmp.len1=tmp.len0=0;        for(int i=st;i<=ed;i++)        {            if(save[i]==0)            {                tmp.len0++;                if(save[i+1]==1)                {                    a.push_back(tmp);                    tmp.len0=tmp.len1=0;                }            }            else            {                tmp.len1++;            }        }        /*        for(int i=0;i<a.size();i++)        {            printf("%d %d\n",a[i].len1,a[i].len0);        }        */        sta.push(a[0]);        node now;        for(int i=1;i<a.size();i++)        {            while(!sta.empty())            {                now=sta.top();                if(a[i].len1*(now.len1+now.len0)>now.len1*(a[i].len1+a[i].len0))                {                    sta.push(a[i]);                    break;                }                else                {                    sta.pop();                    a[i].len1+=now.len1;                    a[i].len0+=now.len0;                    if(sta.empty())                    {                        sta.push(a[i]);                        break;                    }                }            }        }        double ans=0;        while(!sta.empty())        {            now=sta.top();            sta.pop();            double x=double(now.len1)/double(now.len1+now.len0);            ans+=double(now.len1)*(x-1)*(x-1)+double(now.len0)*x*x;        }        //ans=ans/double(n);        printf("%.6f\n",ans);    }    return 0;}

0 0