线段树

线段树就是区间树，在一段区间求最值等问题。
题目：poj3264：
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
简单翻译就是在一段区间求最差：
下面代码：附加解释,代码参考了一下其他人

//这是有关线段树上面的//当然是参考了其他人的代码才完成的，还不行，需要在努力//干巴爹#include <iostream>#include <cstdio>using namespace std; int MAX = -0x3fffffff; int MIN = 0x3fffffff;/*此处用树状数组*/struct Node{    int L,R;    int max,min;    Node *left,*right;};int Rmax(int a,int b){    if(a > b) return a;    return b;}int Rmin(int a,int b){    if(a < b) return a;    return b;}Node Tree[100001];int Count = 0;//建树过程，实际用数组代替void BuildTree(Node *root,int L,int R){    root->L = L,root->R = R;    root->max = MAX;    root->min = MIN;    if(L != R)    {        Count++;        root->left = Tree+Count;        Count++;        root->right = Tree+Count;        BuildTree(root->left,L,(L+R)/2);        BuildTree(root->right,(L+R)/2+1,R);    }}//插入过程，将每个叶子节点插入对应的数值，同时将每个区间的最大值，最小值表示出来void Insert(Node *root,int i,int h){    if(root->L == i && root->R == i)    {        root->max = Rmax(root->max,h);        root->min = Rmin(root->min,h);        return;    }    root->max = Rmax(root->max,h);    root->min = Rmin(root->min,h);    if(i <= (root->L+root->R)/2)        Insert(root->left,i,h);    else        Insert(root->right,i,h);}//查询过程void Quiry(Node *root,int l,int r){    if(root->max <= MAX && root->min >= MIN)        return;    if(root->L == l && root->R == r)    {        MAX = Rmax(MAX,root->max);        MIN = Rmin(MIN,root->min);        //更新最值        return;    }    else if(r <= (root->L+root->R)/2)    {        Quiry(root->left,l,r);    }    else if(l > (root->L+root->R)/2)    {        Quiry(root->right,l,r);    }    else    {        Quiry(root->left,l,(root->L+root->R)/2);        Quiry(root->right,(root->L+root->R)/2+1,r);    }}int main(){    int m,n,h;    scanf("%d%d",&m,&n);    BuildTree(Tree,1,m);    for(int i=1;i<=m;i++)    {        scanf("%d",&h);        Insert(Tree,i,h);    }    while(n--)    {        int l,r;        scanf("%d%d",&l,&r);        MAX = -0x3fffffff;        MIN = 0x3fffffff;        Quiry(Tree,l,r);        printf("%d\n",MAX-MIN);    }}