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  • POJ 3264 Balanced Lineup | st表

    题意:

    求区间max-min

    st表模板

     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<cstring>
     4 #include<cmath>
     5 #define N 50010
     6 using namespace std;
     7 int rmin[N][21],rmax[N][21],n,l,r,q;
     8 int read()
     9 {
    10     int ret=0,neg=1;
    11     char j=getchar();
    12     for (;j>'9' || j<'0';j=getchar())
    13     if (j=='-') neg=-1;
    14     for (;j<='9' && j>='0';j=getchar())
    15     ret=ret*10+j-'0';
    16     return ret*neg;
    17 }
    18 int main()
    19 {
    20     n=read(),q=read();
    21     for (int i=1,x;i<=n;i++)
    22     x=read(),rmin[i][0]=x,rmax[i][0]=x;
    23     for (int j=1;j<=20;j++)
    24     for (int i=1;i+(1<<j)-1<=n;i++)
    25         rmin[i][j]=min(rmin[i][j-1],rmin[i+(1<<j-1)][j-1]),rmax[i][j]=max(rmax[i][j-1],rmax[i+(1<<j-1)][j-1]);
    26     while (q--)
    27     {
    28     l=read(),r=read();
    29     int lg=log2(r-l+1);
    30     printf("%d
    ",max(rmax[l][lg],rmax[r-(1<<lg)+1][lg]) - min(rmin[l][lg],rmin[r-(1<<lg)+1][lg]));
    31     }
    32     return 0;
    33 }
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  • 原文地址:https://www.cnblogs.com/mrsheep/p/7880308.html
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