jzoj 1278_排队_线段树

题目描述

　每天,农夫 John 的N(1 <= N <= 50,000)头牛总是按同一序列排队. 有一天, John决定让一些牛们玩一场飞盘比赛. 他准备找一群在对列中为置连续的牛来进行比赛.但是为了避免水平悬殊,牛的身高不应该相差太大.
　　John 准备了Q (1 <= Q <= 180,000) 个可能的牛的选择和所有牛的身高 (1 <=身高 <= 1,000,000). 他想知道每一组里面最高和最低的牛的身高差别.

---

思路

一个区间求最值的问题，很明显用线段树就可以水过了

---

#include <stdio.h>#define maxn 50001int a[maxn];int ma(int x, int y){    return x > y;}int mi(int x, int y){    return x < y;}struct tree{    int x, min, max;}e[maxn * 4];int tt = 0, maxx = 0, minn = 0x7fffffff;int read(){    int x = 0; char ch = getchar();    while (ch < '0' || ch > '9') ch = getchar();    while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch -'0'; ch = getchar();}    return x;}int make(int p, int l, int r){    if (l == r)     {        tt++;        e[p].x = e[p].max = e[p].min = a[tt];        return 0;    }       int m = (l + r) >> 1;    make(p << 1, l, m);    make(p << 1 | 1, m + 1, r);    e[p].max = e[p << 1].max > e[p << 1 | 1].max ? e[p << 1].max : e[p << 1 | 1].max;    e[p].min = e[p << 1].min < e[p << 1 | 1].min ? e[p << 1].min : e[p << 1 | 1].min;}int count(int p, int l, int r, int x, int y){    if (l == x && r == y)    {        maxx = maxx > e[p].max ? maxx : e[p].max;        minn = minn < e[p].min ? minn : e[p].min;        return 0;    }    int m = (l + r) >> 1;    if (y <= m) count(p << 1, l, m, x ,y);    else if (m < x) count(p << 1 | 1, m + 1, r, x, y);      else    {        count(p << 1, l, m, x, m);        count(p << 1 | 1, m + 1, r, m + 1, y);    }}int main(){    int n, m;    scanf("%d%d", &n, &m);    for (int i = 1; i <= n; i++)        a[i] = read();    make(1, 1, n);    for (int i = 1; i <= m; i++)    {        int x = read(), y = read();        count(1, 1, n, x, y);        printf("%d\n", maxx - minn);        maxx = 0;         minn = 0x7fffffff;    }}