标签：rmq 线段树
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Language:
Balanced Lineup
Description For the daily milking, Farmer John‘s N cows (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 of Q (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, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive. 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 3 1 7 3 4 2 5 1 5 4 6 2 2 Sample Output 6 3 0 Source |
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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-8
typedef __int64 ll;
using namespace std;
#define N 50005
int 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_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int 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-8
typedef __int64 ll;
using namespace std;
#define N 50005
int 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_JUDGE
int 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;
}
poj 3264 Balanced Lineup（简单线段树 或 rmq）
标签：rmq 线段树
原文地址：http://blog.csdn.net/u014737310/article/details/43951597
