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Language: Default 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 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 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;
}