hdu 5033 Building 2014 ACM/ICPC Asia Regional Beijing Online

这一题WA了好久，对拍才查出来错，因为放入单调栈的时候少考虑了前两种元素(1,10),(2,13)这种case，忘了把第二个放进去。

维护一个单调栈(凸包)，两个方向各遍历一次。每次只遍历单调栈，可构成凸包截止。

#include<iostream>#include<stdio.h>#include<cstdio>#include<stdlib.h>#include<vector>#include<string>#include<cstring>#include<cmath>#include<algorithm>#include<stack>#include<queue>#include <ctype.h>using namespace std;int T;int N;int Q;const int maxn=200100;pair<int,int>a[maxn];//first is x,second is hpair<int,int> s[maxn];int top;//top of stack s,指向下一个代存位置pair<int,int> l[maxn];pair<int,int> r[maxn];int q[maxn];//record the location of queryconst double pi=acos(-1.0);bool cmp(pair<int,int>a,pair<int,int>b){    return a.first<b.first;}void input(){    memset(a,0,sizeof(a));    memset(l,0,sizeof(l));    memset(r,0,sizeof(r));    memset(q,0,sizeof(q));    scanf("%d",&N);    for(int i=0;i<N;i++)    {        scanf("%d %d",&a[i].first,&a[i].second);    }    scanf("%d",&Q);    for(int i=N;i<N+Q;i++)    {        scanf("%d",&a[i].first);        q[i-N]=a[i].first;        a[i].second=-(i-N+1);// if<0, then it represent the order of query    }    sort(a,a+N+Q,cmp);//    for(int i=0;i<N+Q;i++)//    {//        cout<<a[i].first<<" "<<a[i].second<<endl;//    }}void convex(int x){    pair<int,int>tmp=s[top]; //   cout<<top<<endl;    while(top>0)    {        tmp=s[top-1];//the first element        if(tmp.second<a[x].second)//need =?        {            top--;//to keep descending        }        else        {            break;        }    }  //  s[top++]=a[x];    while(top>=2)//at least two elements    {        tmp=s[top-1];        double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);        double slope2=1.0*(tmp.second-a[x].second)/abs(a[x].first-tmp.first);        if(slope1>=slope2)        {            top--;        }        else        {            s[top++]=a[x];           // cout<<x<<" xx "<<tmp.second<<" "<<a[x].second<<endl;         //   cout<<x<<" "<<s[top-1].first<<endl;            return;        }    }    if(top==1||top==0)    {        s[top++]=a[x];      //  cout<<x<<"xx "<<s[top-1].second<<endl;   //   cout<<x<<" "<<s[top-1].first<<endl;        return;    }}void run(){    memset(s,0,sizeof(s));  //  memset(l,0,sizeof(l)); //   memset(r,0,sizeof(r));    top=0;    s[top++]=a[0];    for(int i=1;i<N+Q;i++)    {        //cout<<s[top-1].first<<endl;        if(a[i].second>0)        {            convex(i);        }        else        {            while(top>=2)            {                pair<int,int>tmp=s[top-1];                double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);                double slope2=1.0*tmp.second/abs(a[i].first-tmp.first);               //  cout<<slope1<<"  "<<slope2<<endl;                if(slope1>=slope2)                {                    top--;                }                else                {                    break;                }            }            int qry=-a[i].second-1;            l[qry]=s[top-1];         //   cout<<l[qry].first<<endl;;        }    }    memset(s,0,sizeof(s));//from right end   // cout<<"right"<<endl;    top=0;    s[top++]=a[N+Q-1];    for(int i=N+Q-2;i>=0;i--)    {        if(a[i].second>0)        {            convex(i);            //cout<<s[top-1].first<<endl;        }        else        {            while(top>=2)            {                pair<int,int>tmp=s[top-1];                double slope1=1.0*(s[top-2].second-tmp.second)/abs(tmp.first-s[top-2].first);                double slope2=1.0*tmp.second/abs(a[i].first-tmp.first);              //  cout<<tmp.first<<" tmp "<<tmp.second<<endl;                if(slope1>=slope2)                {                    top--;                }                else                {                    break;                }            }            int qry=-a[i].second-1;            r[qry]=s[top-1];            //  cout<<top-1<<" "<<qry<<" "<<r[qry].first<<endl;        }    }}double cal(int x){  //  cout<<x<<" "<<l[x].second<<" "<<q[x]<<" "<<l[x].first<<endl;    // cout<<x<<" "<<r[x].second<<" "<<q[x]<<" "<<r[x].first<<endl;    double theta1=atan2(l[x].second,(q[x]-l[x].first));    double theta2=atan2(r[x].second,(r[x].first-q[x]));    return (pi-theta1-theta2)/pi*180.0;}int main(){   // freopen("input.txt","r",stdin);    freopen("data.txt","r",stdin);  //  freopen("out1.txt","w",stdout);   scanf("%d",&T);   for(int ca=1;ca<=T;ca++)   {       input();       run();       printf("Case #%d:\n",ca);       for(int i=0;i<Q;i++)       {           printf("%.10lf\n",cal(i));       }   }    return 0;}