标签：balance res game each potential ring signed set while
Balanced Lineup
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.
6 3 1 7 3 4 2 5 1 5 4 6 2 2Sample Output
6 3 0
题意:问你[l r]区间最大值－最小值多少
记得写过 今天给学弟写 自己无聊 写了一遍
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<cstdlib> 6 #include<string.h> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<cmath> 13 typedef long long ll; 14 typedef unsigned long long LL; 15 using namespace std; 16 const double PI=acos(-1.0); 17 const double eps=0.0000000001; 18 const int N=100000+10; 19 const int INF=1e9; 20 int a[N]; 21 struct node{ 22 int l,r; 23 int maxx,minn; 24 int val; 25 }tree[N*4]; 26 void build(int left,int right,int pos){ 27 tree[pos].l=left; 28 tree[pos].r=right; 29 tree[pos].val=tree[pos].maxx=0; 30 tree[pos].minn=INF; 31 int mid=(tree[pos].l+tree[pos].r)>>1; 32 if(tree[pos].l==tree[pos].r){ 33 return ; 34 } 35 build(left,mid,pos<<1); 36 build(mid+1,right,pos<<1|1); 37 } 38 void update(int left,int right,int pos,int x,int y){ 39 int mid=(left+right)>>1; 40 if(tree[pos].l==x&&tree[pos].r==x){ 41 tree[pos].val=y; 42 tree[pos].minn=y; 43 tree[pos].maxx=y; 44 return; 45 } 46 if(x>mid)update(mid+1,right,pos<<1|1,x,y); 47 else 48 update(left,mid,pos<<1,x,y); 49 tree[pos].maxx=max(tree[pos<<1].maxx,tree[pos<<1|1].maxx); 50 tree[pos].minn=min(tree[pos<<1].minn,tree[pos<<1|1].minn); 51 return ; 52 53 } 54 int maxx; 55 int minn; 56 int query(int pos,int x,int y){ 57 if(x==tree[pos].l&&tree[pos].r==y){ 58 //cout<<4<<endl; 59 //cout<<tree[pos].maxx<<" "<<tree[pos].minn<<endl; 60 minn=min(minn,tree[pos].minn); 61 maxx=max(maxx,tree[pos].maxx); 62 return 0; 63 } 64 int mid=(tree[pos].l+tree[pos].r)>>1; 65 if(x>tree[pos].r||y<tree[pos].l)return 0; 66 else if(x>mid)query(pos<<1|1,x,y); 67 else if(y<=mid)query(pos<<1,x,y); 68 else{ 69 query(pos<<1|1,mid+1,y); 70 query(pos<<1,x,mid); 71 } 72 return 0; 73 } 74 int main(){ 75 int n,q; 76 scanf("%d%d",&n,&q); 77 build(1,n,1); 78 for(int i=1;i<=n;i++){ 79 scanf("%d",&a[i]); 80 update(1,n,1,i,a[i]); 81 } 82 while(q--){ 83 maxx=0; 84 minn=INF; 85 //cout<<minn<<" "<<maxx<<endl; 86 int x,y; 87 scanf("%d%d",&x,&y); 88 query(1,x,y); 89 cout<<maxx-minn<<endl; 90 } 91 }
POJ 3264 Balanced Lineup (线段树）
标签：balance res game each potential ring signed set while
原文地址：http://www.cnblogs.com/Aa1039510121/p/7134253.html
