POJ 3264

题目链接：http://poj.org/problem?id=3264

——————————————————————————————————————————

题目简述：题意：给出一个长度为N<=10^6的数字串和M<=10^5个操作。操作有一种：求出[L, R]范围中，最大值与最小值的差值

——————————————————————————————————————————

题目关键字：线段树

——————————————————————————————————————————

题目思路：简单的线段树题目。其中find返回值是tree型（因为返回的是一个类型，无所谓局部变量的问题）

——————————————————————————————————————————

源代码：

#include<stdio.h>struct tree{    int max;    int min;    int left;    int right;       }tree[3000000];int n = 0;int cow[500010];int mmax(int a,int b){   if(a>b)  return a;   else return b;    }int mmin(int a,int b){   if(a>b)  return b;   else return a;}void init(int cur,int l,int r){     int m = (l+r)/2;        tree[cur].left = l;    tree[cur].right = r;        if(l == r)    {        tree[cur].max = cow[l];        tree[cur].min = cow[l];        }    else    {        init(cur*2,l,m);        init(cur*2+1,m+1,r);                tree[cur].max = mmax(tree[cur*2].max,tree[cur*2+1].max);        tree[cur].min = mmin(tree[cur*2].min,tree[cur*2+1].min);    }}struct tree find(int cur,int a,int b){     struct tree l,r,ans;          if(a == tree[cur].left && b == tree[cur].right)        return tree[cur];             else     {        int m = (tree[cur].left+tree[cur].right)/2;               if(a>m) return find(cur*2+1,a,b);        else if(b<=m) return find(cur*2,a,b);             else              {                  l = find(cur*2,a,m);                  r = find(cur*2+1,m+1,b);                  ans.max = mmax(l.max,r.max);                  ans.min = mmin(l.min,r.min);                  return ans;             }     }         }int main(){    int i = 0,q = 0;    int a = 0,b = 0;    struct tree ans;        scanf("%d%d",&n,&q);    for(i = 1;i<=n;i++)      scanf("%d",&cow[i]);        init(1,1,n);        for(i = 0;i<q;i++)    {       scanf("%d%d",&a,&b);       ans = find(1,a,b);       printf("%d\n",ans.max-ans.min);    }    return 0;}