題目大意:輸入N個數和Q個查詢,每次查詢區間[L,D]中的最大數和最小數的差。
題解:用線段樹記錄區間的最大值和最小值,輸出相減的結果即可。
#include <iostream>
#include <stdio.h>
using namespace std;
struct node{
int l,r;
int minN,maxN;
};
node tree[200005];
int n,q,a[50005],l,r,maxN,minN;
void buildTree(int root,int l,int r){//建樹
tree[root].l = l;
tree[root].r = r;
if(l == r){
tree[root].minN = a[l];
tree[root].maxN = a[l];
return;
}
buildTree(2*root,l,(l+r)/2);
buildTree(2*root+1,(l+r)/2+1,r);
tree[root].minN = min(tree[2*root].minN,tree[2*root+1].minN);
tree[root].maxN = max(tree[2*root].maxN,tree[2*root+1].maxN);
}
void query(int root,int l,int r){//查詢區間l,r的最大值、最小值,將值存在全局變量
if(tree[root].l == l && tree[root].r == r){
minN = min(minN,tree[root].minN);
maxN = max(maxN,tree[root].maxN);
return;
}
int mid = (tree[root].l+tree[root].r)/2;
if(r <= mid){
query(2*root,l,r);
}
else if(l > mid){
query(2*root+1,l,r);
}
else{
query(2*root,l,mid);
query(2*root+1,mid+1,r);
}
}
int main()
{
scanf("%d%d",&n,&q);
for(int i = 1; i <= n; i ++){
scanf("%d",&a[i]);
}
buildTree(1,1,n);
for(int i = 0; i < q; i ++){
scanf("%d%d",&l,&r);
maxN = 0;
minN = 10000000;
query(1,l,r);
printf("%d\n",maxN - minN);
}
return 0;
}