Balanced Lineup
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
题目类型：线段树
算法分析：线段树的直接应用，每一个节点动态维护区间的两个端点和最小值minval最大值maxval。递归建树的过程中将叶节点的最大值和最小值进行初始化。然后对于每一个区间查询，使用两个全局变量maxval和minval来记录在查找的区间中的最大值和最小值。最后直接输出两者之差即可
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#include <iostream> #include <fstream> #include <algorithm> #include <iomanip> #include <cstring> #include <cstdio> #include <cmath> #include <map> #include <string> #include <vector> #include <stack> #include <queue> #include <set> #include <list> #include <ctime> using namespace std; const int root = 1; const int INF = 66666666; const int maxn = 50000 + 66; int maxval, minval; struct Node { int L, R; int maxval, minval; }; Node tree[maxn*3]; void BuildTree (int pos, int L, int R) { tree[pos].L = L; tree[pos].R = R; if (L == R) { scanf ("%I64d", &tree[pos].maxval); tree[pos].minval = tree[pos].maxval; return ; } int mid = (L + R) >> 1; BuildTree (pos * 2, L, mid); BuildTree (pos * 2 + 1, mid + 1, R); tree[pos].minval = min (tree[pos*2].minval, tree[pos*2+1].minval); tree[pos].maxval = max (tree[pos*2].maxval, tree[pos*2+1].maxval); } void Query (int pos, int L, int R) { int mid = (tree[pos].L + tree[pos].R) >> 1; if (L == tree[pos].L && R == tree[pos].R) { maxval = max (maxval, tree[pos].maxval); minval = min (minval, tree[pos].minval); } else if (R <= mid) Query (pos << 1, L, R); else if (L > mid) Query ((pos << 1) + 1, L, R); else { Query (pos << 1, L, mid); Query ((pos << 1) + 1, mid + 1, R); } } /* void Update (int pos, int k, int add) { if (k >= tree[pos].L && k <= tree[pos].R) { tree[pos].sum += add; if (tree[pos].L == tree[pos].R) return ; Update (pos * 2, k, add); Update (pos * 2 + 1, k, add); } return ; }*/ int main() { // freopen ("aaa.txt", "r", stdin); int n, q; scanf ("%d%d", &n, &q); BuildTree (root, 1, n); int i, temp_a, temp_b; for (i = 0; i < q; i++) { scanf ("%d%d", &temp_a, &temp_b); maxval = -INF, minval = INF; Query (root, temp_a, temp_b); printf ("%d\n", maxval - minval); } return 0; } |
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