poj3264 Balanced Lineup(线段树)
2015-01-19 20:54
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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
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 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
Sample Output
Source
USACO 2007 January Silver
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 36480 | Accepted: 17081 | |
Case Time Limit: 2000MS |
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 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
USACO 2007 January Silver
/* 注意不能直接返回根节点的最大值减去最小值。。。。加油!! Time:2015-1-19 20:54 */ #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long LL; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 const int MAX=50000+10; #define INF 0x3f3f3f3f LL minV,maxV; struct Tree{ LL minV,maxV; }tree[MAX<<2]; void Build(int l,int r,int rt){ tree[rt].maxV=-INF; tree[rt].minV=INF; if(l==r)return; int mid=(l+r)>>1; Build(lson); Build(rson); } void Insert(int pos,LL val,int l,int r,int rt){ if(pos==l&&pos==r){ tree[rt].minV=val; tree[rt].maxV=val; return; } int mid=(l+r)>>1; if(pos<=mid)Insert(pos,val,lson); else if(pos>mid)Insert(pos,val,rson); tree[rt].maxV=max(tree[rt].maxV,val); tree[rt].minV=min(tree[rt].minV,val); } void Query(int L,int R,int l,int r,int rt){ if(L==l&&R==r){ minV=min(minV,tree[rt].minV); maxV=max(maxV,tree[rt].maxV); return; } int mid=(l+r)>>1; if(R<=mid){ Query(L,R,lson); }else if(L>mid){ Query(L,R,rson); }else{ Query(L,mid,lson); Query(mid+1,R,rson); } } int main(){ int n,Q; LL val; scanf("%d%d",&n,&Q); memset(tree,0,sizeof(tree)); Build(1,n,1); for(int i=1;i<=n;i++){ scanf("%lld",&val); Insert(i,val,1,n,1); } int L,R; //printf("%d %d\n",tree[1].maxV,tree[1].minV); while(Q--){ minV=INF;maxV=-INF; scanf("%d%d",&L,&R); Query(L,R,1,n,1);//不能直接返回tree[rt].maxV-tree[rt].minV, printf("%lld\n",maxV-minV); } return 0; }
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