ST表 || RMQ问题 || BZOJ 1699: [Usaco2007 Jan]Balanced Lineup排队 || Luogu P2880 [USACO07JAN]平衡的阵容Balanced Lineup
题面：P2880 [USACO07JAN]平衡的阵容Balanced Lineup
题解：
ST表板子
代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #define max(a,b) ((a)>(b)?(a):(b)) 5 #define min(a,b) ((a)<(b)?(a):(b)) 6 using namespace std; 7 const int maxn=50050,max_log=18,maxlog=17,inf=1<<30; 8 int Fmx[maxn][max_log],N,Q,Fmn[maxn][max_log],LOG[maxn],A,B; 9 int main(){ 10 scanf("%d%d",&N,&Q); 11 for(int i=0;i<=N;i++) 12 for(int j=0;j<=maxlog;j++) 13 Fmn[i][j]=inf; 14 for(int i=1;i<=N;i++){ 15 scanf("%d",&Fmx[i][0]); 16 Fmn[i][0]=Fmx[i][0]; 17 } 18 LOG[0]=-1; 19 for(int i=1;i<=N;i++)LOG[i]=LOG[i>>1]+1; 20 for(int j=1;j<=maxlog;j++)// 21 for(int i=1;i+(1<<j)-1<=N;i++){ 22 Fmx[i][j]=max(Fmx[i][j-1],Fmx[i+(1<<(j-1))][j-1]); 23 Fmn[i][j]=min(Fmn[i][j-1],Fmn[i+(1<<(j-1))][j-1]); 24 } 25 while(Q--){ 26 scanf("%d%d",&A,&B); 27 int x=LOG[B-A+1],ans1,ans2; 28 ans1=max(Fmx[A][x],Fmx[B-(1<<x)+1][x]); 29 ans2=min(Fmn[A][x],Fmn[B-(1<<x)+1][x]); 30 printf("%d\n",ans1-ans2); 31 } 32 return 0; 33 }
By:AlenaNuna
