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POJ 3264 Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 33094 | Accepted: 15552 | |
| 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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
打算用两种方法做。顺便差别一下，RMQ和线段树的差别。他们都都说RMQ比线段树好，我发现时间也差不了多少，尽管都没优化
AC代码例如以下：
线段树。
///线段树 2250MS 2404K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define M 50010
#define inf 100000000
using namespace std;
struct H
{
int l,r,maxx,minn;
}trees[4*M];
int n,m;
int num[M];
void build_trees(int jd ,int l,int r)
{
trees[jd].l=l;trees[jd].r=r;
if(l==r)
{
trees[jd].maxx=num[l];
trees[jd].minn=num[l];
return ;
}
int mid = (l+r)/2;
build_trees(jd*2,l,mid);
build_trees(jd*2+1,mid+1,r);
trees[jd].maxx=max(trees[jd*2].maxx,trees[jd*2+1].maxx);
trees[jd].minn=min(trees[jd*2].minn,trees[jd*2+1].minn);
}
int query_max(int jd,int l,int r)
{
int ans=0;
if(l<=trees[jd].l&&r>=trees[jd].r)
return trees[jd].maxx;
int mid = (trees[jd].l+trees[jd].r)/2;
if(l<=mid) ans=max(ans,query_max(jd*2,l,r)) ;
if(r>mid) ans=max(ans,query_max(jd*2+1,l,r));
return ans;
}
int query_min(int jd,int l,int r)
{
int ans=inf;
if(l<=trees[jd].l&&r>=trees[jd].r)
return trees[jd].minn;
int mid = (trees[jd].l+trees[jd].r)/2;
if(l<=mid) ans=min(ans,query_min(jd*2,l,r)) ;
if(r>mid) ans=min(ans,query_min(jd*2+1,l,r));
return ans;
}
int main()
{
int i,j;
int a,b;
while(~scanf("%d%d",&n,&m))
{
memset(num,0,sizeof num);
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
build_trees(1,1,n);
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
printf("%d\n",query_max(1,a,b)-query_min(1,a,b));
}
}
return 0;
}
RMQ！。！
///RMQ 1813MS 12100K
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define M 50010
#define inf 100000000
using namespace std;
int n,m;
int num[M];
int dp1[M][30],dp2[M][30];
void RMQ_min()
{
int i,j;
memset(dp1,0,sizeof dp1);
for(i=1;i<=n;i++)
dp1[i][0]=num[i];
for(j=1;1<<j<=n;j++)
for(i=1;i+(1<<j)-1<=n;i++)
dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
}
void RMQ_max()
{
int i,j;
memset(dp2,0,sizeof dp2);
for(i=1;i<=n;i++)
dp2[i][0]=num[i];
for(j=1;1<<j<=n;j++)
for(i=1;i+(1<<j)-1<=n;i++)
dp2[i][j]=max(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
}
int rmq_min(int l,int r)
{
int i,j;
int k=0;
while(1<<(k+1)<=r-l+1)
k++;
return min(dp1[l][k],dp1[r-(1<<k)+1][k]);
}
int rmq_max(int l,int r)
{
int i,j;
int k=0;
while(1<<(k+1)<=r-l+1)
k++;
return max(dp2[l][k],dp2[r-(1<<k)+1][k]);
}
int main()
{
int i,j;
int a,b;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
RMQ_min();
RMQ_max();
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
printf("%d\n",rmq_max(a,b)-rmq_min(a,b));
}
}
return 0;
}POJ 3264 Balanced Lineup