POJ_3264_Balance Lineup_RMQ

八十道POJ了，加油！

题意：

线段上每个点有一个值，多次询问某些区间最大值和最小值的差。

裸的sparse-table解决RMQ，当然用线段树也一样，RMQ一定注意边界情况想清楚，否则样例一水WA都不知道怎么WA的。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>using namespace std;#define mxn 200010#define logmxn 50int maxx[mxn][logmxn],minn[mxn][logmxn];int h[mxn];int n,q;void set(){for(int i=0;i<n;++i)maxx[i][0]=minn[i][0]=h[i];for(int j=1;(1<<j)<=n;++j)for(int i=0;i+(1<<j)<=n;++i){maxx[i][j]=max(maxx[i][j-1],maxx[i+(1<<(j-1))][j-1]);minn[i][j]=min(minn[i][j-1],minn[i+(1<<(j-1))][j-1]);}}int ans(int l,int r){int k=0;while(l+(1<<(k+1))<=r+1)++k;return max(maxx[l][k],maxx[r-(1<<k)+1][k])-min(minn[l][k],minn[r-(1<<k)+1][k]);}int main(){while(scanf("%d%d",&n,&q)!=EOF){for(int i=0;i<n;++i)scanf("%d",&h[i]);set();int l,r;for(int i=0;i<q;++i){scanf("%d%d",&l,&r);printf("%d\n",ans(l-1,r-1));}}return 0;}