UVa 12299 线段树 单点更新 RMQ with Shifts

因为shift操作中的数不多，所以直接用单点更新模拟一下就好了。

太久不写线段树，手好生啊，不是这错一下就是那错一下。

PS：输入写的我有点蛋疼，不知道谁有没有更好的写法。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 100000 + 10;
const int maxnode = (maxn << 2);
const int INF = 0x3f3f3f3f;

void scan(int& x)
{
    x = 0;
    char c = ' ';
    while(c < '0' || c > '9') c = getchar();
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
}

int qL, qR;
int p, v;
int n, Q;
int minv[maxnode];

int a[maxn];

void build(int o, int L, int R)
{
    if(L == R) { scan(a[L]); minv[o] = a[L]; return ; }
    int M = (L + R) / 2;
    build(o<<1, L, M);
    build(o<<1|1, M+1, R);
    minv[o] = min(minv[o<<1], minv[o<<1|1]);
}

void update(int o, int L, int R)
{
    //if(v >= minv[o]) return ;
    if(L == R) { minv[o] = v; return ; }

    int M = (L + R) / 2;
    if(p <= M) update(o<<1, L, M);
    else update(o<<1|1, M+1, R);
    minv[o] = min(minv[o<<1], minv[o<<1|1]);
}

int query(int o, int L, int R)
{
    if(qL <= L && R <= qR) return minv[o];
    int ans = INF;
    int M = (L + R) / 2;
    if(qL <= M) ans = min(ans, query(o<<1, L, M));
    if(qR > M) ans = min(ans, query(o<<1|1, M+1, R));
    return ans;
}

char op[50];

int main()
{
    while(scanf("%d%d", &n, &Q) == 2 && n)
    {
        build(1, 1, n);
        while(Q--)
        {
            scanf("%s", op);
            int l = strlen(op);
            if(op[0] == 'q')
            {
                int i = 6;
                qL = 0;
                while(i < l && op[i] >= '0' && op[i] <= '9') { qL = qL * 10 + op[i] - '0'; i++; }
                while(i < l && (op[i] < '0' || op[i] > '9')) i++;
                qR = 0;
                while(i < l && op[i] >= '0' && op[i] <= '9') { qR = qR * 10 + op[i] - '0'; i++; }
                printf("%d
", query(1, 1, n));
            }
            else
            {
                vector<int> hehe;

                for(int i = 6; i < l;)
                {
                    while(i < l && (op[i] < '0' || op[i] > '9')) i++;
                    if(i >= l) break;
                    int x = 0;
                    while(i < l && op[i] >= '0' && op[i] <= '9') { x = x * 10 + op[i] - '0'; i++; }
                    hehe.push_back(x);
                }

                int sz = hehe.size();
                for(int i = 0; i < sz - 1; i++)
                {
                    v = a[hehe[i + 1]];
                    p = hehe[i];
                    update(1, 1, n);
                }
                v = a[hehe[0]];
                p = hehe[sz - 1];
                update(1, 1, n);

                int t = a[hehe[0]];
                for(int i = 0; i < sz - 1; i++) a[hehe[i]] = a[hehe[i+1]];
                a[hehe[sz - 1]] = t;
            }
        }
    }

    return 0;
}