poj3264

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 23031		Accepted: 10702
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

线段树。

#include<iostream>
#include<algorithm>
using namespace std;
#define MY_MAX -99999999
#define MY_MIN 999999999

struct CNode {
    int L, R;
    int nMin, nMax;
    CNode * pLeft, * pRight;
};
int nMax, nMin;
CNode Tree[1000000]; //其实两倍叶子节点的数量+1就够 int nCount = 0; //总节点数目
int nCount = 0;

void BuildTree(CNode * pRoot, int L, int R) {
    pRoot->L = L;
    pRoot->R = R;
    pRoot->nMin = MY_MIN;
    pRoot->nMax = MY_MAX;
    if (L != R) {
        nCount++;
        pRoot->pLeft = Tree + nCount;
        nCount++;
        pRoot->pRight = Tree + nCount;
        BuildTree(pRoot->pLeft, L, (L + R) / 2);
        BuildTree(pRoot->pRight, (L + R) / 2 + 1, R);
    }
}

void Insert(CNode * pRoot, int i, int v) {
    //将第i个数,其值为v,插入线段树 {
    if (pRoot->L == i && pRoot->R == i) {
        pRoot->nMin = pRoot->nMax = v;
        return;
    }
    pRoot->nMin = min(pRoot->nMin, v);
    pRoot->nMax = max(pRoot->nMax, v);
    if (i <= (pRoot->L + pRoot->R) / 2)
        Insert(pRoot->pLeft, i, v);
    else
        Insert(pRoot->pRight, i, v);
}

void Query(CNode * pRoot, 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 {
    if (pRoot->nMin >= nMin && pRoot->nMax <= nMax) return;
    if (s == pRoot->L && e == pRoot->R) {
        nMin = min(pRoot->nMin, nMin);
        nMax = max(pRoot->nMax, nMax);
        return;
    }
    if (e <= (pRoot->L + pRoot->R) / 2)
        Query(pRoot->pLeft, s, e);
    else if (s >= (pRoot->L + pRoot->R) / 2 + 1)
        Query(pRoot->pRight, s, e);
    else {
        Query(pRoot->pLeft, s, (pRoot->L + pRoot->R) / 2);
        Query(pRoot->pRight, (pRoot->L + pRoot->R) / 2 + 1, e);
    }
}

int main() {
    int n, q, h;
    int i, j, k;
    scanf("%d%d", &n, &q);
    nCount = 0;
    BuildTree(Tree, 1, n);
    for (i = 1; i <= n; i++) {
        scanf("%d", &h);
        Insert(Tree, i, h);
    }
    for (i = 0; i < q; i++) {
        int s, e;
        scanf("%d%d", &s, &e);
        nMax = MY_MAX;
        nMin = MY_MIN;
        Query(Tree, s, e);
        printf("%d\n", nMax - nMin);
    }
    return 0;
}