Lightoj1082——Array Queries（线段树+求区间最小值）

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
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).

Output
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
2

5 3
78 1 22 12 3
1 2
3 5
4 4

1 1
10
1 1
Output for Sample Input
Case 1:
1
3
12
Case 2:
10

很裸的线段树

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 25#define Mod 10001using namespace std;const int maxn=222222;int min(int a,int b){    if(a<b)        return a;    else        return b;}int MIN[maxn<<2];void PushUp(int rt){    MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);}void build(int l,int r,int rt){    if(r==l)    {        scanf("%d",&MIN[rt]);        return;    }    int m=(l+r)>>1;    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);    PushUp(rt);}int query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)        return MIN[rt];    int m=(l+r)>>1;    int ret=INF;    if(L<=m)        ret=min(ret,query(L,R,l,m,rt<<1));    if(R>m)        ret=min(ret,query(L,R,m+1,r,rt<<1|1));    return ret;}int main(){    int t;    scanf("%d",&t);    for(int cas=1; cas<=t; ++cas)    {        int n,q,l,r;        scanf("%d%d",&n,&q);        build(1,n,1);        printf("Case %d:\n",cas);        while(q--)        {            scanf("%d%d",&l,&r);            printf("%d\n",query(l,r,1,n,1));        }    }    return 0;}