Developing School's Contest 2012-8 by HVCST problem -- 1010

weak node

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Problem Description

Country C preparing to attack Country J, but the shortage of troops Country C, Country C to find out the location of weak Country J defense.

Input

Line 1: n m(1<=n<=100,000, 1<=m<=120,000 )
Line 2: n numbers, respectively said the defense force of the Country J n nodes
Line 3 to 2+m: m action, expressed as follows
  find:(i,j) 
  From i to j nodes to identify the weakest defense of the node, and the output of Defense
  move:(i1,i2.....)
  These points will move relative to each other position
  For example, if [8,7,6,5,4,3,2,1], then move(3,5,7) yields [8,7,4,5,2,3,6,1]

Output

For each find, print the weakest defense of the value

Sample Input

8 3
8 7 6 5 4 3 2 1
find(2,5)
move(3,5,7)
find(2,5)

Sample Output

4
2

题目意思：给定一个数组，有两种指令
find指令包含两个数a和b，求区间a到b间最小的值
move指令包含不确定的数，让数组的按这些数循环左移一个位置
这题和去年湖南程序大赛K题一样

采用线段树处理：
find指令直接输出区间最小值
move通过处理更新线段树

//Accepted    1010    203 MS    1672 KB

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 100001;

int st_min[maxn<<2];
int A[maxn];
inline int minn(int a,int b) { return a>b?b:a; }

void PushUP(int rt)
{
    st_min[rt] = minn(st_min[rt<<1],st_min[rt<<1|1]);

}
void build(int l,int r,int rt) {
    if (l == r)
    {
        st_min[rt] = A[l] ;
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUP(rt);
}
void update(int p,int add,int l,int r,int rt) {
    if (l == r) 
    {
        st_min[rt] = add;
        return ;
    }
    int m = (l + r) >> 1;
    if (p <= m) update(p , add , lson);
    else update(p , add , rson);
    PushUP(rt);
}
int query_min(int L,int R,int l,int r,int rt) 
{
    if (L <= l && r <= R) 
    {
        return st_min[rt];
    }
    int m = (l + r) >> 1;
    int ret1 = 0x7fffffff,ret2 = 0x7fffffff;
    if (L <= m) ret1 = query_min(L , R , lson);
    if (R > m) ret2 = query_min(L , R , rson);
    return minn(ret1,ret2);
}

int k[maxn];
char op[100];

int main(void)
{
    int i,n,a,b,m,t;
    scanf("%d%d",&n,&m);    
    for(i=1;i<=n;i++)
        scanf("%d",&A[i]);
    build(1 , n , 1);
    while(m--)
    {
        scanf("%s",op);
        if(op[0]=='f')
        {
            sscanf(op,"find(%d,%d)",&a,&b);
            printf("%d\n",query_min(a,b,1,n,1));
        }
        else
        {

            char *p = strtok(op + 5, "," );
            int x = 0;
            while( p )
            {
                sscanf( p, "%d", &k[x] );
                p = strtok( NULL, "," );
                x++;
            }
            t = A[k[0]];
            for(i=0;i<x-1;i++)
            {
                A[k[i]] = A[k[i+1]];
                update(k[i],A[k[i+1]],1,n,1);
            }
            A[k[x-1]] = t;
            update(k[x-1],t,1,n,1);
                
        }
    }
    return 0;
}