【BZOJ】1699 [Usaco2007 Jan]Balanced Lineup排队

【算法】线段树

#include<cstdio>
#include<cctype>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f,maxn=50010;
struct tree{int l,r,mins,maxs;}t[maxn*3];
int n,q,a[maxn];
int read_t;char c;
int read()
{
    read_t=0;
    while(!isdigit(c=getchar()));
    do{read_t=read_t*10+c-'0';}while(isdigit(c=getchar()));
    return read_t;
}
void build(int k,int l,int r)
{
    t[k].l=l;t[k].r=r;
    if(l==r)t[k].mins=a[l],t[k].maxs=a[r];
     else
      {
          int mid=(l+r)>>1;
          build(k<<1,l,mid);
          build(k<<1|1,mid+1,r);
          t[k].mins=min(t[k<<1].mins,t[k<<1|1].mins);
          t[k].maxs=max(t[k<<1].maxs,t[k<<1|1].maxs);
      }
}
int askmaxs(int k,int l,int r)
{
    int left=t[k].l,right=t[k].r;
    if(l<=left&&r>=right)return t[k].maxs;
     else
      {
          int mid=(left+right)>>1,ans=0;
          if(l<=mid)ans=askmaxs(k<<1,l,r);
          if(r>mid)ans=max(ans,askmaxs(k<<1|1,l,r));
          return ans;
      }
}
int askmins(int k,int l,int r)
{
    int left=t[k].l,right=t[k].r;
    if(l<=left&&r>=right)return t[k].mins;
     else
      {
          int mid=(left+right)>>1,ans=inf;
          if(l<=mid)ans=askmins(k<<1,l,r);
          if(r>mid)ans=min(ans,askmins(k<<1|1,l,r));
          return ans;
      }
}
int main()
{
    n=read();q=read();
    for(int i=1;i<=n;i++)a[i]=read();
    build(1,1,n);
    for(int i=1;i<=q;i++)
     {
         int u=read(),v=read();
         printf("%d\n",askmaxs(1,u,v)-askmins(1,u,v));
     }
    return 0;
}
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posted @ 2016-12-19 21:07  ONION_CYC  阅读(143)  评论(0编辑  收藏  举报