nyoj 568——RMQ with Shifts——————【线段树单点更新、区间求最值】

RMQ with Shifts

时间限制： 1000 ms  |  内存限制：65535 KB

难度： 3

 

描述

    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}. 

 

输入

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.

输出

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

样例输入

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) 

样例输出

1 

4 

6 

来源

湖南省第七届大学生计算机程序设计竞赛

题目大意：有个shift操作，也就是常见的更新，只是这里的更新比较别致，是交换i和i+1，i+1和i+2，i+2和i+3，etc（也有人理解为移动i、i+1、i+2...对应的数组中的值）。然后就是询问区间内的最小值。

解题思路：就是按题意一直更新就好了。

#include<bits/stdc++.h>

using namespace std;

#define mid (L+R)/2

#define lson rt*2,L,mid

#define rson rt*2+1,mid+1,R

const int maxn=110000;

const int INF=1e9;

int minv[maxn*4];

int a[maxn],cnt=0;

int b[5000];

char str[100];

void PushUP(int rt){

        minv[rt]=min(minv[rt*2],minv[rt*2+1]);

}

void build(int rt,int L,int R){

    if(L==R){

        scanf("%d",&minv[rt]);

        a[cnt++]=minv[rt];

        return ;

    }

    build(lson);

    build(rson);

    PushUP(rt);

}

int get_b(int st,int en){

    int cnt=0;

    int tm,tmp=0;

    for(int i=st;i<en;i++){

        if(str[i]>='0'&&str[i]<='9'){

           tm=str[i]-'0';

           tmp*=10;

           tmp+=tm;

        }else{

            b[cnt++]=tmp;

            tmp=0;

        }

    }

    return cnt;

}

void exchange(int la,int ra){

    int tm=a[la];

    a[la]=a[ra];

    a[ra]=tm;

}

int query(int rt,int L,int R,int l_ran,int r_ran){

    if(l_ran<=L&&R<=r_ran){

        return minv[rt];

    }

    int ret_l=INF,ret_r=INF;

    if(l_ran<=mid){

        ret_l=query(lson,l_ran,r_ran);

    }

    if(r_ran>mid){

        ret_r=query(rson,l_ran,r_ran);

    }

    return min(ret_l,ret_r);

}

void update(int rt,int L,int R,int pos,int val){

    if(L==R){

        minv[rt]=val;

        return ;

    }

    if(pos<=mid){

        update(lson,pos,val);

    }else{

        update(rson,pos,val);

    }

    PushUP(rt);

}

void debug(){

 for(int i=1;i<16;i++)

    printf("%d %d\n",i,minv[i]);

}

int main(){

    int n,q,m,ans;

    while(scanf("%d%d",&n,&q)!=EOF){

        cnt=0;

        build(1,1,n);

        for(int i=0;i<q;i++){

            scanf("%s",&str);

            int len=strlen(str);

            m= get_b(6,len);

            if(str[0]=='q'){

                ans=query(1,1,n,b[0],b[1]);

                printf("%d\n",ans);

            }else{

                b[m]=b[0];

                for(int i=0;i<m;i++){   //估计瞌睡了，这里迷糊了好久

                    update(1,1,n,b[i],a[b[i+1]-1]);

                }

                for(int i=0;i<m-1;i++){

                    exchange(b[i]-1,b[i+1]-1);

                }

            }

        }

    }

    return 0;

}