此题为入门级线段树
题意:给定Q(1<=Q<=200000)个数A1A2…AQ,多次求任一区间Ai-Aj中最大数和最小数的差
#include<algorithm>
#include<cstdio>
#include<string>
#include<string.h>
#include<iostream>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e3 + 10; int minV = INF;
int maxV = -INF;
struct Node
{
int L, R;
int minV, maxV;
//Node *pLeft, *pRight;
int Mid()
{
return (L + R) / 2;
}
};
Node tree[800010];//4倍叶子节点的数量就够 void BuildTree(int root, int L, int R)
{
tree[root].L = L;
tree[root].R = R;
tree[root].minV = INF;
tree[root].maxV = -INF;
if (L != R)
{
BuildTree(2 * root + 1, L, (L + R) / 2);
BuildTree(2 * root + 2, (L + R) / 2 + 1, R);
}
} void Insert(int root, int i, int v)
//将第i个数,其值为v,插入线段树
{
if (tree[root].L == tree[root].R)
{
//成立则亦有tree[root].R==i
tree[root].minV = tree[root].maxV = v;
return;
}
tree[root].minV = min(tree[root].minV, v);
tree[root].maxV = max(tree[root].maxV, v);
if (i <= tree[root].Mid())
Insert(2 * root + 1, i, v);
else
Insert(2 * root + 2, i, v);
} void Query(int root, 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 (tree[root].minV >= minV&&tree[root].maxV <= maxV)
return;
if (tree[root].L == s&&tree[root].R == e)
{
minV = min(minV, tree[root].minV);
maxV = max(maxV, tree[root].maxV);
return;
}
if (e <= tree[root].Mid())
Query(2 * root + 1, s, e);
else if (s > tree[root].Mid())
Query(2 * root + 2, s, e);
else
{
Query(2 * root + 1, s, tree[root].Mid());
Query(2 * root + 2, tree[root].Mid() + 1, e);
}
} int main()
{
int n, q, h;
int i, j, k;
scanf("%d%d", &n, &q);
BuildTree(0, 1, n);
for (i = 1; i <= n; i++)
{
scanf("%d", &h);
Insert(0, i, h);
}
for (i = 0; i < q; i++)
{
int s, e;
scanf("%d%d", &s, &e);
minV = INF;
maxV = -INF;
Query(0, s, e);
printf("%d\n", maxV - minV);
}
return 0;
}