Beta Round ＃9 (酱油杯noi考后欢乐赛)乌鸦喝水

题目：http://www.contesthunter.org/contest/Beta%20Round%20%EF%BC%839%20%28%E9%85%B1%E6%B2%B9%E6%9D%AFnoi%E8%80%83%E5%90%8E%E6%AC%A2%E4%B9%90%E8%B5%9B%29/%E4%B9%8C%E9%B8%A6%E5%96%9D%E6%B0%B4

题解：真是一道神题！考场上绝对想不到标算orz！

题解写在代码注释里

代码：

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 100000+5
 26 
 27 #define maxm 500+100
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 54 
 55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 int n,m,h,s[maxn],a[maxn],b[maxn],id[maxn];
 61 inline void add(int x,int y){for(;x<=n;x+=x&(-x))s[x]+=y;}
 62 inline int sum(int x){int t=0;for(;x;x-=x&(-x))t+=s[x];return t;}
 63 inline bool cmp(int x,int y){return b[x]==b[y]?x<y:b[x]<b[y];}
 64 
 65 int main()
 66 
 67 {
 68 
 69     freopen("input","r",stdin);
 70 
 71     freopen("output.txt","w",stdout);
 72 
 73     n=read();m=read();h=read();
 74     for1(i,n)a[i]=read();
 75     for1(i,n)
 76     {
 77         int x=read();
 78         if(a[i]<=h)b[i]=(h-a[i])/x+1;//b[i]表示该点最多可以被饮几次水
 79         if(b[i]>0)add(i,1);id[i]=i;
 80     }
 81     //可以证明，最后的答案一定是某个b[i]，所以我们按b[i]从小到大处理
 82     sort(id+1,id+n+1,cmp);
 83     int now=1,pos=0,ans=0,st=n+1;//now 表示当前的趟数
 84     //for1(i,n)cout<<i<<' '<<id[i]<<' '<<b[i]<<endl;
 85     for1(i,n)if(b[id[i]]){st=i;break;}
 86     for2(i,st,n)
 87     {
 88         while(now<=m)
 89         {
 90             int t=sum(n)-sum(pos);
 91             if(ans+t<b[id[i]])ans+=t,now++,pos=0;//如果可以走完这一趟
 92             else break;
 93         }
 94         if(now>m)break;
 95         int l=pos,r=n,t=sum(pos);
 96         while(l<=r)
 97         {
 98             int mid=(l+r)>>1;
 99             if(sum(mid)-t+ans>=b[id[i]])r=mid-1;else l=mid+1;
100         }//二分到下一个点
101         ans=b[id[i]];pos=l;
102         for(;b[id[i]]==b[id[i+1]];i++)add(id[i],-1);//去除不能再次饮水的点
103         add(id[i],-1);
104     }
105     printf("%d
",ans);
106 
107     return 0;
108 
109 }

View Code