kuangbin专题七 POJ3264 Balanced Lineup （线段树最大最小）-白红宇

kuangbin专题七 POJ3264 Balanced Lineup （线段树最大最小）

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发布时间：2019-06-24

本文共 3005 字，大约阅读时间需要 10 分钟。

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

Lines 2..

N+1: Line

i+1 contains a single integer that is the height of cow

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 31734251 54 62 2

Sample Output

630 线段树维护最大最小，不涉及更改，只用pushup query就可以了

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 #include 
         
           6 #include 
          
            7 #include 
           
             8 #include 
            
              9 #include 
             10 #include 
              
               11 #include 
               
                12 #include 
                
                 13 #include 
                 
                  14 using namespace std;15 #define FO freopen("in.txt","r",stdin);16 #define rep(i,a,n) for (int i=a;i
                  
                   =a;i--)18 #define pb push_back19 #define mp make_pair20 #define all(x) (x).begin(),(x).end()21 #define fi first22 #define se second23 #define SZ(x) ((int)(x).size())24 #define debug(x) cout << "&&" << x << "&&" << endl;25 #define lowbit(x) (x&-x)26 #define mem(a,b) memset(a, b, sizeof(a));27 typedef vector
                   
                     VI;28 typedef long long ll;29 typedef pair
                    
                      PII;30 const ll mod=1000000007;31 const int inf = 0x3f3f3f3f;32 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}34 //head35 36 const int maxn=200010;37 int minn[maxn<<2],maxx[maxn<<2],n,q,a[maxn],maxpos,minpos;38 39 void pushup(int rt) {40 minn[rt]=min(minn[rt<<1],minn[rt<<1|1]);41 maxx[rt]=max(maxx[rt<<1],maxx[rt<<1|1]);42 }43 44 void build(int rt,int L,int R){45 minn[rt]=0;46 maxx[rt]=0;47 if(L==R) {48 scanf("%d",&a[rt]);49 minn[rt]=maxx[rt]=a[rt];50 return;51 }52 int mid=(L+R)>>1;53 build(rt<<1,L,mid);54 build(rt<<1|1,mid+1,R);55 pushup(rt);56 }57 58 void query(int rt,int L,int R,int l,int r) {59 if(L>=l&&R<=r) {60 minpos=min(minpos,minn[rt]);61 maxpos=max(maxpos,maxx[rt]);62 return;63 }64 int mid=(L+R)>>1;65 if(l<=mid) query(rt<<1,L,mid,l,r);66 if(r>mid) query(rt<<1|1,mid+1,R,l,r);67 }68 69 int main() {70 while(~scanf("%d%d",&n,&q)) {71 build(1,1,n);72 int l,r;73 while(q--) {74 maxpos=-1,minpos=inf;75 scanf("%d%d",&l,&r);76 query(1,1,n,l,r);77 printf("%d\n",l==r?0:maxpos-minpos);78 }79 }80 }

转载于:https://www.cnblogs.com/ACMerszl/p/9865238.html

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