标签：des style blog http color strong
链接：http://poj.org/problem?id=3264
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 32772 | Accepted: 15421 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
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思路也很清晰，就是重新弄一个root[]数组求最小值，开始把minn设为0，找了好久才找到
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <algorithm> #include <iostream> using namespace std; #define Maxx 50010 int str[Maxx],root[Maxx<<2],root1[Maxx<<2]; int n; int minn,maxx; void make_tree(int l,int r,int rt) { if(l == r) { root[rt]=str[l]; root1[rt]=str[l]; return ; } int mid=(l+r)/2; make_tree(l,mid,rt*2); make_tree(mid+1,r,rt*2+1); root[rt]=max(root[rt*2],root[rt*2+1]); root1[rt]=min(root1[rt*2],root1[rt*2+1]); } void update(int l,int r,int rt,int a,int b) { if(l == r && l == a) { root[rt]=b; root1[rt]=b; return ; } int mid=(l+r)/2; if(a<=mid) update(l,mid,rt*2,a,b); else update(mid+1,r,rt*2+1,a,b); root[rt]=max(root[rt*2],root[rt*2+1]); root1[rt]=min(root1[rt*2],root1[rt*2+1]); } void query(int l,int r,int rt,int left,int right) { if(l == left&&r == right) { maxx=max(maxx,root[rt]); minn=min(minn,root1[rt]); return ; } int mid=(l+r)/2; if(left>mid) { query(mid+1,r,rt*2+1,left,right); } else if(right<=mid) { query(l,mid,rt*2,left,right); } else { query(l,mid,rt*2,left,mid); query(mid+1,r,rt*2+1,mid+1,right); } } int main() { int q; int i,j; int a,b; while(scanf("%d%d",&n,&q)!=EOF) { for(i=1;i<=n;i++) { scanf("%d",&str[i]); } make_tree(1,n,1); for(i=1;i<=q;i++) { minn=100000000;maxx=0; scanf("%d%d",&a,&b); query(1,n,1,a,b); printf("%d\n",maxx-minn); } } return 0; }
poj 3264 Balanced Lineup(线段数求区间最大最小值),布布扣,bubuko.com
poj 3264 Balanced Lineup(线段数求区间最大最小值)
标签：des style blog http color strong
原文地址：http://www.cnblogs.com/ccccnzb/p/3841135.html
