Balanced Lineup
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 62103 Accepted: 29005
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
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
【代码】：
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int minV=INF;
int maxV=-INF;
struct Node
{
int L,R;//区间起点和终点
int minV,maxV;//本区间里的最大最小值
int Mid(){
return (L+R)/2;
}
};
Node tree[800010];
void BuildTree(int root, int L, int R)
{
tree[root].L = L;
tree[root].R = R;
tree[root].minV = INF;
tree[root].maxV = -INF;
if(L != R)
{
BuildTree(2*root+1, L, (L+R)/2);
BuildTree(2*root+2, (L+R)/2 + 1, R);
}
}
void Insert(int root, int i ,int v)//将第i个数，其值为v，插入线段树
{
if(tree[root].L == tree[root].R)
{
tree[root].minV = tree[root].maxV = v;
return ;
}
tree[root].minV = min(tree[root].minV, v);
tree[root].maxV = max(tree[root].maxV, v);
if(i <= tree[root].Mid())
Insert(2*root+1,i,v);
else
Insert(2*root+2,i,v);
}
void Query(int root, int s, int e)//查询区间[s,e]中的最小值和最大值，如果更优就记在全局变量里
//minV和maxV里
{
if(tree[root].minV >= minV && tree[root].maxV <= maxV)
return;
if(tree[root].L == s && tree[root].R == e)
{
minV = min(minV, tree[root].minV);
maxV = max(maxV, tree[root].maxV);
return;
}
if(e <= tree[root].Mid())
Query(2*root+1, s, e);
else if(s > tree[root].Mid())
Query(2*root+2, s, e);
else
{
Query(2*root+1, s, tree[root].Mid());
Query(2*root+2, tree[root].Mid()+1, e);
}
}
int main()
{
int n,q,h;
int i,j,k;
scanf("%d%d",&n,&q);
BuildTree(0,1,n);
for(i=1;i<=n;i++)
{
scanf("%d",&h);
Insert(0,i,h);
}
for(i=0;i<q;i++)
{
int s,e;
scanf("%d%d",&s,&e);
minV = INF;
maxV = -INF;
Query(0,s,e);
printf("%d\n",maxV - minV);
}
}
/*
6 3
1 7 3 4 2 5
1 5
4 6
2 2
*/
