| Time Limit: 3 second(s) | Memory Limit: 64 MB |
Given an array with N elements, indexed from 1 to N. Now you will be given some queries in the form I J, your task is to find the minimum value from index I to J.
Input starts with an integer T (≤ 5), denoting the number of test cases.
The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 105), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers range in [0, 105].
The next q lines will contain a query which is in the form I J (1 ≤ I ≤ J ≤ N).
For each test case, print the case number in a single line. Then for each query you have to print a line containing the minimum value between index I and J.
Sample Input |
Output for Sample Input |
|
2
5 3 78 1 22 12 3 1 2 3 5 4 4
1 1 10 1 1 |
Case 1: 1 3 12 Case 2: 10 |
Dataset is huge. Use faster I/O methods.
线段树，区间最值问题。
AC代码：
#include <iostream> #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; const int maxn = 200010; int Max[maxn<<2]; void pushup(int root) { Max[root]=min(Max[root<<1],Max[root<<1|1]); } void build(int root,int L,int R) { if(L==R){ scanf("%d",&Max[root]); return; } int mid=(L+R)>>1; build(root<<1,L,mid); build(root<<1|1,mid+1,R); pushup(root); } void update(int root,int L,int R,int i,int v) { if(L==R){ Max[root]=v; return; } int mid=(L+R)>>1; if(i<=mid)update(root<<1,L,mid,i,v); else update(root<<1|1,mid+1,R,i,v); pushup(root); } int query(int root,int L,int R,int s,int e) { if(s==L&&e==R) { return Max[root]; } int mid=(L+R)>>1; int res=INF; if(e<=mid) res=min(res,query(root<<1,L,mid,s,e)); else if(s>mid) res=min(res,query(root<<1|1,mid+1,R,s,e)); else { res=min(res,query(root<<1,L,mid,s,mid)); res=min(res,query(root<<1|1,mid+1,R,mid+1,e)); } return res; } int main() { int N,M,a,b; int t; scanf("%d",&t); int xp=0; while(t--){ scanf("%d%d",&N,&M); build(1,1,N); printf("Case %d:\n",++xp); while(M--){ scanf("%d%d",&a,&b); int ans=query(1,1,N,a,b); printf("%d\n",ans); } } return 0; }
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LightOJ Array Queries 1082【线段树求区间最值】
原文地址：http://blog.csdn.net/ydd97/article/details/47816331
