对区间建树,维护区间最大值最小值。
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 43493 | Accepted: 20419 | |
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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
Source
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <map> #include <set> #include <stack> #include <cmath> #include <string> #include <iostream> #include <vector> #define LL long long #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 using namespace std; const int maxn=1<<30; const int SIZE=5e4+10; //int x[15]={0,31,30,31,30,31,30,31,31,30,31,30,31}; struct seg{ int high,low; }a[SIZE<<2]; void pushup(int rt){ a[rt].high=max(a[rt<<1].high,a[rt<<1|1].high); a[rt].low=min(a[rt<<1].low,a[rt<<1|1].low); } void build(int l,int r,int rt){ if(l==r){ int x; scanf("%d",&x); a[rt].low=a[rt].high=x; return; } int m=(l+r)>>1; build(lson); build(rson); pushup(rt); } void query(int &high,int &low,int L,int R,int l,int r,int rt){ if(L<=l&&r<=R){ if(a[rt].high>high)high=a[rt].high; if(a[rt].low<low)low=a[rt].low; return ; } int m=(l+r)>>1; if(L<=m)query(high,low,L,R,lson); if(R>m)query(high,low,L,R,rson); } int main() { int n,q,x,y; while(scanf("%d%d",&n,&q)!=EOF){ build(1,n,1); for(int i=0;i<q;i++){ scanf("%d%d",&x,&y); int high=-maxn,low=maxn; query(high,low,x,y,1,n,1); printf("%d\n",high-low); } } return 0; }