题目大意:
有N个点,Q次查询,每次查询区间内的最大值和最小值之差
思路: 线段树
代码:
#include <iostream> #include<cstdio> #include<cstring> using namespace std; const int MAX = 200000+5; int N,Q,a[MAX]; int ansMin,ansMax; struct Tree { int l,r,mins,maxs; }tree[4*MAX]; ///建立线段树,cnt为节点角标,a[l]-a[r]建树 void build(int cnt,int l,int r) { tree[cnt].l = l,tree[cnt].r = r; if(l==r) { tree[cnt].mins = tree[cnt].maxs = a[l]; return; } int mid = (l + r)/2; build(2*cnt,l,mid); build(2*cnt+1,mid+1,r); ///回溯建立父节点 tree[cnt].mins = min(tree[2*cnt].mins,tree[2*cnt+1].mins); tree[cnt].maxs = max(tree[2*cnt].maxs,tree[2*cnt+1].maxs); } ///查询区间l-r的最大值和最小值 void query(int cnt,int l,int r) { if(tree[cnt].mins>=ansMin&&tree[cnt].maxs<=ansMax) return; if(tree[cnt].l==l&&tree[cnt].r==r) { ansMax = max(tree[cnt].maxs,ansMax), ansMin = min(tree[cnt].mins,ansMin); return; } ///分三种情况讨论 int mid = (tree[cnt].l+tree[cnt].r)/2; if(r<=mid)///在mid右端 query(2*cnt,l,r); else if(l>mid)///在mid左端 query(2*cnt+1,l,r); else///两端 { query(2*cnt,l,mid); query(2*cnt+1,mid+1,r); } } int main() { //freopen("in.txt","r",stdin); while(scanf("%d%d",&N,&Q)!=EOF) { memset(tree,0,sizeof(tree)); for(int i=1;i<=N;i++) scanf("%d",&a[i]); int left,right; build(1,1,N); for(int i=1;i<=Q;i++) { ansMin = 0x3f3f3f3f, ansMax = -1; scanf("%d%d",&left,&right); query(1,left,right); printf("%d\n",ansMax-ansMin); } } return 0; }
时间: 2024-08-11 07:42:34