Language:
Balanced Lineup
Time Limit:5000MSMemory Limit:65536K
Total Submissions:36833Accepted:17252
Case Time Limit:2000MS
Description
For the daily milking, Farmer John’sNcows (1 ≤N≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list ofQ(1 ≤Q≤ 200,000) potential groups of cows and their heights (1 ≤height≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Line 1: Two space-separated integers,NandQ.Lines 2..N+1: Linei+1 contains a single integer that is the height of cowiLinesN+2..N+Q+1: Two integersAandB(1 ≤A≤B≤N), representing the range of cows fromAtoBinclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 31734251 54 62 2
Sample Output
630
Source
求区间最大减最小
线段树代码：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;using namespace std;#define N 50005int n,m,a[N];struct stud{ int le,ri; int mi,ma;}f[N*4];void build(int pos,int le,int ri){f[pos].le=le;f[pos].ri=ri;if(le==ri){f[pos].mi=f[pos].ma=a[le];return ;}int mid=MID(le,ri);build(L(pos),le,mid);build(R(pos),mid+1,ri);f[pos].ma=max(f[L(pos)].ma,f[R(pos)].ma);f[pos].mi=min(f[L(pos)].mi,f[R(pos)].mi);}int querymin(int pos,int le,int ri){if(f[pos].le>=le&&f[pos].ri<=ri)return f[pos].mi; int mid=MID(f[pos].le,f[pos].ri); if(mid>=ri)return querymin(L(pos),le,ri); if(mid<le)return querymin(R(pos),le,ri); return min(querymin(L(pos),le,mid),querymin(R(pos),mid+1,ri));}int querymax(int pos,int le,int ri){if(f[pos].le>=le&&f[pos].ri<=ri)return f[pos].ma; int mid=MID(f[pos].le,f[pos].ri); if(mid>=ri)return querymax(L(pos),le,ri); if(mid<le)return querymax(R(pos),le,ri); return max(querymax(L(pos),le,mid),querymax(R(pos),mid+1,ri));}int main(){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);#endif // ONLINE_JUDGEint i,j;while(~scanf("%d%d",&n,&m)){for(i=1;i<=n;i++)scanf("%d",&a[i]);build(1,1,n);int le,ri;while(m–){scanf("%d%d",&le,&ri);printf("%d\n",querymax(1,le,ri)-querymin(1,le,ri));}}return 0;}
RMQ 代码：
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8typedef __int64 ll;using namespace std;#define N 50005int dpmin[N][20];int dpmax[N][20];int n,m,a[N];void inint(){int i,j;for(i=1;i<=n;i++)dpmin[i][0]=dpmax[i][0]=a[i];for(j=1;(1<<j)<=n+1;j++)for(i=1;i+(1<<j)-1<=n;i++){dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]);dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]);}}inline int getmax(int le,int ri){int k=(int)(log(ri-le+1+0.0)/log(2.0));return max(dpmax[le][k],dpmax[ri-(1<<k)+1][k]);}inline int getmin(int le,int ri){int k=(int)(log(ri-le+1+0.0)/log(2.0));return min(dpmin[le][k],dpmin[ri-(1<<k)+1][k]);}int main(){//#ifndef ONLINE_JUDGE// freopen("in.txt","r",stdin);//#endif // ONLINE_JUDGEint i,j;while(~scanf("%d%d",&n,&m)){for(i=1;i<=n;i++)scanf("%d",&a[i]);inint();int le,ri;while(m–){scanf("%d%d",&le,&ri);printf("%d\n",getmax(le,ri)-getmin(le,ri));}}return 0;}
，你并不一定会从此拥有更美好的人生，
