Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 27962 | Accepted: 13136 | |
| Case Time Limit: 2000MS | ||
Description
For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range
of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest
cow in the group.
Input
Line 1: Two space-separated integers, N andQ.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤B ≤
N), representing the range of cows from A toB inclusive.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤B ≤
N), representing the range of cows from A toB inclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
本题是个简单的线段树区间查询
#include <iostream> #include<cstdio> #include <algorithm> #include <numeric> using namespace std; #define MY_MIN 99999999 #define MY_MAX -99999999 struct CNode { int L,R; int nMin,nMax; CNode * pLeft, * pRight; }; int nMax, nMin; CNode Tree[1000000];//其实两倍叶子节点的数量+1就够 int nCount = 0; ////总节点数目 void BuildTree(CNode * pRoot, int L,int R) { pRoot->L = L; pRoot->R = R; pRoot->nMin = MY_MIN; pRoot->nMax = MY_MAX; if ( L != R) { nCount ++; pRoot->pLeft = Tree + nCount; nCount ++; pRoot->pRight = Tree + nCount; BuildTree( pRoot->pLeft, L, ( L + R )/2); BuildTree( pRoot->pRight, (L + R) / 2 + 1,R); } } void Insert( CNode * pRoot , int i,int v) //将第i个数，其值为v，插入线段树 { if( pRoot->L == i && pRoot->R == i ) { pRoot->nMin = pRoot->nMax = v; return; } pRoot->nMin = min(pRoot->nMin,v); pRoot->nMax =max(pRoot->nMax,v); if( i <= (pRoot->L + pRoot->R )/2 ) Insert( pRoot->pLeft,i,v); else Insert( pRoot->pRight,i,v); } void Query(CNode * pRoot, 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( pRoot->nMin >= nMin && pRoot->nMax <= nMax ) return; if( s == pRoot->L && e == pRoot->R) { nMin = min(pRoot->nMin,nMin); nMax =max(pRoot->nMax,nMax); return ; } if( e <= (pRoot->L + pRoot->R) / 2 ) Query(pRoot->pLeft, s,e); else if( s >= (pRoot->L + pRoot->R) / 2 + 1) Query(pRoot->pRight, s,e); else { Query(pRoot->pLeft, s,(pRoot->L + pRoot->R) / 2); Query(pRoot->pRight, (pRoot->L + pRoot->R) / 2+1 ,e); } } int main() { int n,q,h; int i,j,k; scanf("%d%d",&n,&q); nCount = 0; BuildTree(Tree,1,n); for( i = 1;i <= n;i ++ ) { scanf("%d",&h); Insert( Tree,i,h); } for( i = 0;i < q;i ++ ) { int s,e; scanf("%d%d", &s,&e); nMax = MY_MAX; nMin = MY_MIN; Query(Tree,s,e); printf("%d\n",nMax - nMin); } return 0; }