poj 3264线段树

题意：  给出一个数列，  找出一个区间中的最大值与最小值， 并求它们的差。

思路： 简单线段树。

AC代码：

View Code 
  1 #include <cstdio>
  2 #include <iostream>
  3 #include <string>
  4 #include <cstring>
  5 #include <queue>
  6 using namespace std;
  7 const int N = 50100;
  8 struct NODE
  9 {
 10     int left, right, min_v, max_v;
 11 }node[N*4];
 12 
 13 int n, q, value[N];
 14 
 15 void build(int left, int right, int pos)
 16 {
 17     node[pos].left = left;
 18     node[pos].right = right;
 19     if(left == right)
 20     {
 21         node[pos].max_v = node[pos].min_v = value[left];
 22         return ;
 23     }
 24 
 25     int mid = (left+right)/2;
 26     build(left, mid, pos*2);
 27     build(mid+1, right, pos*2+1);
 28     node[pos].max_v = max(node[pos*2].max_v, node[pos*2+1].max_v);
 29     node[pos].min_v = min(node[pos*2].min_v, node[pos*2+1].min_v);
 30 }
 31 
 32 int findmax(int l, int r, int pos)
 33 {
 34     if(l == node[pos].left && r == node[pos].right)
 35     {
 36         return node[pos].max_v;
 37     }
 38 
 39     int mid = (node[pos].left+node[pos].right)/2;
 40 
 41     if(r <= mid)
 42     {
 43         return findmax(l, r, 2*pos);
 44     }
 45     else if(l > mid)
 46     {
 47         return findmax(l, r, 2*pos+1);
 48     }
 49     else
 50     {
 51         return max( findmax(l,mid, 2*pos), findmax(mid+1, r, 2*pos+1) );
 52     }
 53 }
 54 
 55 int findmin(int l, int r, int pos)
 56 {
 57     if(l == node[pos].left && r == node[pos].right)
 58     {
 59         return node[pos].min_v;
 60     }
 61 
 62     int mid = (node[pos].left+node[pos].right)/2;
 63 
 64     if(r <= mid)
 65     {
 66         return findmin(l, r, 2*pos);
 67     }
 68     else if(l > mid)
 69     {
 70         return findmin(l, r, 2*pos+1);
 71     }
 72     else
 73     {
 74         return min( findmin(l,mid, 2*pos), findmin(mid+1, r, 2*pos+1) );
 75     }
 76 }
 77 
 78 int main()
 79 {
 80     int a, b, max_v, min_v;
 81     while(scanf("%d%d", &n, &q) != EOF)
 82     {
 83         for(int i=1; i<=n; i++)
 84         {
 85             scanf("%d", &value[i]);
 86         }
 87 
 88         build(1, n, 1);
 89 
 90         while(q--)
 91         {
 92             scanf("%d%d", &a, &b);
 93             max_v = findmax(a, b, 1);
 94             min_v = findmin(a, b, 1);
 95            //cout<<max_v<<"  "<<min_v<<endl;
 96             printf("%d\n", max_v - min_v);
 97         }
 98     }
 99     return 0;
100 }
posted @ 2012-12-02 20:21  Gu Feiyang  阅读(165)  评论(0)    收藏  举报