Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 75294 | Accepted: 34483 | |
| Case Time Limit: 2000MS | ||
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
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
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
The meaning of problems
The difference between the maximum and minimum values of the query section
solve
A solution: ST table
After pretreatment by O (n * log (n) ) in the query a value within the interval most O (1) time.
The nature of the pretreatment is DP, define an array dp dp [i] [j], dp [i] [j] represents: starting from the i-th, j successive values of 2 ^ most number can be easily obtained: dp [i] [0] is the i-th itself
Then transfer equation can be obtained: DP [I] [J] = max (DP [I] [J-. 1], DP [I ^ 2 + (. 1-J)] [-J. 1])
Then we can find an array of value dp all positions by preprocessing
When the query, the first calculating section length required for 2 logarithm, i.e. obtaining the above equation j, then the equation O (1) to the query
Solution two: segment tree
Code
ST table
1 #include <iostream> 2 #include <algorithm> 3 #include <cmath> 4 #define ll long long 5 #define ull unsigned long long 6 #define ms(a,b) memset(a,b,sizeof(a)) 7 const int inf=0x3f3f3f3f; 8 const ll INF=0x3f3f3f3f3f3f3f3f; 9 const int maxn=1e6+10; 10 const int mod=1e9+7; 11 const int maxm=1e3+10; 12 is the using namespace STD; 13 is // RMQ [i] [J] represents the maximum / minimum values on the first 2 ^ i i bits from the beginning of the 14 // RMQ [i] [J] = max (RMQ [i] [ J-. 1], RMQ [I + (. 1 << (J-. 1))] [J-. 1]) 15 int rmq_max [MAXN] [ 30 ]; 16 int rmq_min [MAXN] [ 30 ]; . 17 int A [MAXN ]; 18 is int main ( int argc, char const * the argv []) . 19 { 20 is #ifndef ONLINE_JUDGE 21 is The freopen ( " /home/wzy/in.txt " ,"r", stdin); 22 freopen("/home/wzy/out.txt", "w", stdout); 23 srand((unsigned int)time(NULL)); 24 #endif 25 ios::sync_with_stdio(false); 26 cin.tie(0); 27 int n,q; 28 cin>>n>>q; 29 for(int i=1;i<=n;i++) 30 cin>>a[i],rmq_min[i][0]=a[i],rmq_max[i][0]=a[i]; 31 for(int j=1;(1<<j)<=n;j++) 32 for(int i=1;i+(1<<(j-1))-1<=n;i++) 33 rmq_min[i][j]=min(rmq_min[i][j-1],rmq_min[i+(1<<(j-1))][j-1]),rmq_max[i][j]=max(rmq_max[i][j-1],rmq_max[i+(1<<(j-1))][j-1]); 34 while(q--) 35 { 36 int x,y; 37 cin>>x>>y; 38 int z=(int)(log(y-x+1)/log(2)); 39 int ans=max(rmq_max[x][z],rmq_max[y-(1<<z)+1][z])-min(rmq_min[x][z],rmq_min[y-(1<<z)+1][z]); 40 cout<<ans<<"\n"; 41 } 42 #ifndef ONLINE_JUDGE 43 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 44 #endif 45 return 0; 46 }
Segment tree
Less likely to tree line, the code is a little ugly
1 #include <iostream> 2 #include <algorithm> 3 #define ll long long 4 #define ull unsigned long long 5 #define ms(a,b) memset(a,b,sizeof(a)) 6 const int inf=0x3f3f3f3f; 7 const ll INF=0x3f3f3f3f3f3f3f3f; 8 const int maxn=1e6+10; 9 const int mod=1e9+7; 10 const int maxm=1e3+10; 11 using namespace std; 12 int a[maxn]; 13 struct wzy 14 { 15 int value_min,value_max; 16 }p[maxn]; 17 inline int lson(int p){return p<<1;} 18 inline int rson(int p){return p<<1|1;} 19 inline void push_max(int o) 20 { 21 p[o].value_max=max(p[lson(o)].value_max,p[rson(o)].value_max); 22 } 23 inline void push_min(int o) 24 { 25 p[o].value_min=min(p[lson(o)].value_min,p[rson(o)].value_min); 26 } 27 void build(int o,int l,int r) 28 { 29 if(l==r) 30 { 31 p[o].value_min=p[o].value_max=a[l]; 32 return ; 33 } 34 int mid=(l+r)>>1; 35 build(lson(o),l,mid); 36 build(rson(o),mid+1,r); 37 push_max(o); 38 push_min(o); 39 } 40 int res,res1; 41 int query_min(int nl,int nr,int l,int r,int o) 42 { 43 if(nl<=l&&nr>=r) 44 return min(res,p[o].value_min); 45 int mid=(l+r)>>1; 46 if(nl<=mid) 47 res=query_min(nl,nr,l,mid,lson(o)); 48 if(nr>mid) 49 res=query_min(nl,nr,mid+1,r,rson(o)); 50 return res; 51 } 52 int query_max(int nl,int nr,int l,int r,int o) 53 { 54 if(nl<=l&&nr>=r) 55 return max(res1,p[o].value_max); 56 int mid=(l+r)>>1; 57 if(nl<=mid) 58 res1=query_max(nl,nr,l,mid,lson(o)); 59 if(nr>mid) 60 res1=query_max(nl,nr,mid+1,r,rson(o)); 61 return res1; 62 } 63 int main(int argc, char const *argv[]) 64 { 65 #ifndef ONLINE_JUDGE 66 freopen("/home/wzy/in.txt", "r", stdin); 67 freopen("/home/wzy/out.txt", "w", stdout); 68 srand((unsigned int)time(NULL)); 69 #endif 70 ios::sync_with_stdio(false); 71 cin.tie(0); 72 int n,q; 73 cin>>n>>q; 74 for(int i=1;i<=n;i++) 75 cin>>a[i]; 76 build(1,1,n); 77 int x,y; 78 while(q--) 79 { 80 cin>>x>>y; 81 res=inf;res1=0; 82 cout<<query_max(x,y,1,n,1)-query_min(x,y,1,n,1)<<"\n"; 83 } 84 #ifndef ONLINE_JUDGE 85 cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; 86 #endif 87 return 0; 88 }