HDU 4923 room and moor greedy + Stack

Link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4923

Bi can be a decimal number.

Idea: a witty idea. For M 1 + N 0 parts in a row, the optimal solution must be bi = m/(m + n ), because Bi is incremental (you can push it manually), if Bi> Bi in the previous section appears in the next section, it is impossible to obtain the optimal solution. Therefore, the two sections are combined for processing, in this case, the stack is needed for maintenance.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <cstdlib>#include <queue>#include <stack>#include <vector>#include <ctype.h>#include <algorithm>#include <string>#include <set>#define PI acos(-1.0)#define maxn 105#define INF 0x7fffffff#define eps 1e-8typedef long long LL;typedef unsigned long long ULL;using namespace std;int aa[1000005];struct Line{    int t0,t1;    double val;    void calc()    {        val=(double)t1/(double)(t1+t0);    }    double sum()    {        return val*val*(double)(t0)+(1-val)*(1-val)*(double)(t1);    }} l[1000005],h;int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(l,0,sizeof(l));        memset(aa,0,sizeof(aa));        int tot,head=-1,tail=-1;        scanf("%d",&tot);        for(int i=0; i<tot; i++)            scanf("%d",&aa[i]);        for(int i=0; i<tot; i++)            if(aa[i]==1)            {                head=i;                break;            }        for(int i=tot-1; i>=0; i--)            if(aa[i]==0)            {                tail=i;                break;            }        if(tail<=head)            printf("0.000000\n");        else        {            int t=head,top=0;            while(t<tail)            {                for(int i=t;i<=tail;i++)                    if(aa[i]==0)                    {                        l[top].t1=i-t;                        t=i;                        break;                    }                for(int i=t;i<=tail;i++)                {                    if(i==tail)                    {                        l[top].t0=i-t+1;                        t=i;                        break;                    }                    else if(aa[i]==1)                    {                        l[top].t0=i-t;                        t=i;                        break;                    }                }                    l[top].calc();                    top++;            }            stack < Line > st;            while(!st.empty())                st.pop();            for(int i=0;i<top;i++)            {                if(st.empty())                    st.push(l[i]);                else                {                    while(!st.empty()&&st.top().val>l[i].val)                    {                        h=st.top();                        st.pop();                        l[i].t0+=h.t0;                        l[i].t1+=h.t1;                        l[i].calc();                    }                    st.push(l[i]);                }            }            double ans=0;            while(!st.empty())            {                h=st.top();                ans+=h.sum();                st.pop();            }            printf("%.6lf\n",ans);        }    }    return 0;}