csu 1110 RMQ with Shifts 线段树

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题意：求动态rmq

思路：因为要移动的数据很小，不过是移动十几个，所以可以每次一边移动后更新

题目链接：http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1110

#include <stdio.h>#include <algorithm>#include <iostream>#include <string.h>using namespace std;#define lson l, m, rt << 1#define rson m+1, r, rt << 1 | 1const int maxn = 111111;int minn[maxn<<2];void pushup(int rt){    minn[rt] = min(minn[rt << 1], minn[rt << 1 | 1]);}void build(int l, int r, int rt){    if(l == r)    {        scanf("%d", &minn[rt]);        return;    }    int m = (l+r) >> 1;    build(lson);    build(rson);    pushup(rt);}int query(int L, int R, int l, int r, int rt){    if(L <= l && R >= r)        return minn[rt];    int m = (l+r) >> 1;    int res = 0x3f3f3f3f;    if(m >= L) res = min(res, query(L, R, lson));    if(m < R) res = min(res, query(L, R, rson));    return res;}void update(int l, int r, int rt, int loc, int val){    if(l == r)    {        minn[rt] = val;        return;    }    int m = (l+r) >> 1;    if(m >= loc) update(lson, loc, val);    else update(rson, loc, val);    pushup(rt);}int main(){    int n, q, cnt;    char s[35];    int a[30];    scanf("%d%d", &n, &q);    build(1, n, 1);    while(q--)    {        scanf("%s", s);        if(s[0] == 'q')        {            int l = 0, r = 0, i;            for(i = 6; s[i] != ','; i++)                l = l*10 + s[i]-'0';            for(i++; s[i] != ')'; i++)                r = r*10 + s[i]-'0';            printf("%d\n", query(l, r, 1, n, 1));        }        else        {            cnt = 0;            int i = 6;            while(s[i] != ')' && s[i] != '\0')//把shift的数存到a数组            {                int num = 0;                while(s[i] != ',' && s[i] != ')')                {                    num = num * 10 + s[i] - '0';                    i++;                }                i++;                a[cnt++] = num;            }            int tmp = query(a[0], a[0], 1, n, 1);//这几行都是在循环移动            for(int i = 0; i < cnt-1; i++)            {                int c = (i+1)%cnt;                int rr = query(a[c], a[c], 1, n, 1);                update(1, n, 1, a[i], rr);            }            update(1, n, 1, a[cnt-1], tmp);        }    }    return 0;}

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