# [USACO07JAN]平衡的阵容Balanced Lineup BZOJ 1699

## 题目描述

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.

## 输入输出格式

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.

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.

## 输入输出样例

6 3
1
7
3
4
2
5
1 5
4 6
2 2

6
3
0线段树维护区间 max min 即可
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 500005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}

ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/

ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}

int n, m;
struct node {
int l, r;
int maxx, minn;
}tree[maxn];

void pushup(int rt) {
tree[rt].maxx = max(tree[rt << 1].maxx, tree[rt << 1 | 1].maxx);
tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}

void build(int l, int r, int rt) {
tree[rt].l = l; tree[rt].r = r;
if (l == r) {
int x; rdint(x); tree[rt].maxx = tree[rt].minn = x;
return;
}
int mid = (l + r) >> 1;
build(l, mid, rt << 1); build(mid + 1, r, rt << 1 | 1);
pushup(rt);
}

int query1(int L, int R,  int rt) {
if (L <= tree[rt].l&&tree[rt].r <= R) {
return tree[rt].maxx;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
int ans = -inf;
if (L <= mid)ans = max(ans, query1(L, R, rt << 1));
if (mid < R)ans = max(ans, query1(L, R, rt << 1 | 1));
return ans;
}

int query2(int L, int R, int rt) {
if (L <= tree[rt].l&&tree[rt].r <= R) {
return tree[rt].minn;
}
int mid = (tree[rt].l + tree[rt].r) >> 1;
int ans = inf;
if (L <= mid)ans = min(ans, query2(L, R, rt << 1));
if (mid < R)ans = min(ans, query2(L, R,  rt << 1 | 1));
return ans;
}

int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m);
build(1, n, 1);
while (m--) {
int a, b; rdint(a); rdint(b);
cout << query1(a, b, 1) - query2(a, b, 1) << endl;
}
return 0;
}


posted @ 2018-11-22 21:02  NKDEWSM  阅读(116)  评论(0编辑  收藏  举报