TOJ 4325 RMQ with Shifts（线段树的单点更新）

 RMQ with Shifts

Time Limit: 1 Sec  Memory Limit: 128 MB  64bit IO Format: %lld
Submitted: 41  Accepted: 18

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Description

In the traditional RMQ (Range Minimum Query) problem, we have a static array A. Then for each query (L, R) (L<=R), we report the minimum value among A[L], A[L+1], …, A[R]. Note that the indices start from 1, i.e. the left-most element is A[1].
In this problem, the array A is no longer static: we need to support another operation shift(i1, i2, i3, …, ik) (i1<i2<...<ik, k>1): we do a left “circular shift” of A[i1], A[i2], …, A[ik].
For example, if A={6, 2, 4, 8, 5, 1, 4}, then shift(2, 4, 5, 7) yields {6, 8, 4, 5, 4, 1, 2}. After that, shift(1,2) yields {8, 6, 4, 5, 4, 1, 2}.

Input

There will be only one test case, beginning with two integers n, q (1<=n<=100,000, 1<=q<=120,000), the number of integers in array A, and the number of operations. The next line contains n positive integers not greater than 100,000, the initial elements in
array A. Each of the next q lines contains an operation. Each operation is formatted as a string having no more than 30 characters, with no space characters inside. All operations are guaranteed to be valid. Warning: The dataset is large, better to use faster
I/O methods.

Output

For each query, print the minimum value (rather than index) in the requested range.

Sample Input 

7 5
6 2 4 8 5 1 4
query(3,7)
shift(2,4,5,7)
query(1,4)
shift(1,2)
query(2,2)

Sample Output

1
4
6

题解：

题目很水，就是题目意思理解了半天，就是询问的话询问区间最小值，shift就是把shift中的坐标全部左移，比如shift（2,4,6）,那么a[4]就到a[2],a[6]就到a[2],a[2]就到a[6],就是一个单点更新加询问的题

代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<deque>
using namespace std;
const int N=100005;
struct node
{
int l,r;
int minn;//区间最小值
}t[N*4];
int a
,d;
void Build(int l,int r,int num)//日常建树
{
t[num].l=l;
t[num].r=r;
if(l==r)
{
scanf("%d",&t[num].minn);
a[d]=t[num].minn;//用一个数组存起来到时候交换就方便了
d++;
return;
}
int mid=(l+r)/2;
Build(l,mid,num*2);
Build(mid+1,r,num*2+1);
t[num].minn=min(t[num*2].minn,t[num*2+1].minn);//更新
}
int query(int l,int r,int num)//日常询问
{
if(l==t[num].l&&r==t[num].r)
{
return t[num].minn;
}
int mid=(t[num].l+t[num].r)/2;
if(r<=mid)
return query(l,r,num*2);
else if(l>mid)
return query(l,r,num*2+1);
else
{
return min(query(l,mid,num*2),query(mid+1,r,num*2+1));
}
}
void update(int pos,int x,int num)//日常更新
{
if(t[num].l==pos&&t[num].r==pos)
{
t[num].minn=x;
return;
}
int mid=(t[num].l+t[num].r)/2;
if(pos<=mid)
update(pos,x,num*2);
else
{
update(pos,x,num*2+1);
}
t[num].minn=min(t[num*2].minn,t[num*2+1].minn);
}
int main()
{
int n,q,i,j,l,r,tag,tot;
d=1;
int b[35];
char s[105];
scanf("%d%d",&n,&q);
Build(1,n,1);//日常建树
for(i=0;i<q;i++)
{
tot=0;
memset(b,0,sizeof(b));//用于保存区间数据
scanf("%s",s);//拆一下字符串
if(s[0]=='q')
tag=0;
else
tag=1;
for(j=0;j<strlen(s);j++)//理解下拆
{
if(s[j]>='0'&&s[j]<='9')
{
b[tot]=b[tot]*10+s[j]-'0';
}
else if(s[j]==','||s[j]==')')
{
tot++;
}
}
if(!tag)//如果是询问
{
printf("%d\n",query(b[0],b[1],1));
}
else//更新
{
update(b[tot-1],a[b[0]],1);//先更新最后一个
for(j=1;j<tot;j++)//前面的覆盖掉后面的
{
update(b[j-1],a[b[j]],1);
}
int temp=a[b[0]];
for(j=0;j<tot-1;j++)//更新下数组
{
a[b[j]]=a[b[j+1]];
}
a[b[tot-1]]=temp;
}
}
}