UVA 12299 - RMQ with Shifts(線段樹)

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UVA 12299 - RMQ with Shifts

題目連結

題意：給定一個數組，兩種操作，每次query操作輸出區間最小值，每次shift操作把選中位置每個位置向左移一位，最左的到最後去

思路：線段樹，shift操作中位置個數不會超過30個，那麼直接當作點修改來做，那麼就變成了簡單的線段樹了

代碼：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define lson(x) ((x<<1) + 1)#define rson(x) ((x<<1) + 2)const int N = 100005;int n, m, shif[35], sn, num[N];char Q[1005];struct Node {    int l, r, Min;} node[4 * N];void handle(char *Q) {    int len = strlen(Q);    int num = -1;    sn = 0;    for (int i = 0; i < len; i++) {if (Q[i] >= '0' && Q[i] <= '9') {    if (num == -1) num = Q[i] - '0';    else num = num * 10 + Q[i] - '0';}else {    if (num != -1) {shif[sn++] = num;num = -1;    }}    }    sort(shif, shif + sn);}void build(int l, int r, int x = 0) {    node[x].l = l; node[x].r = r;    if (l == r) {node[x].Min = num[l];return;    }    int mid = (l + r) / 2;    build(l, mid, lson(x));    build(mid + 1, r, rson(x));    node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min);}void set(int k, int v, int x = 0) {    if (node[x].l == node[x].r) {node[x].Min = v;return;    }    int mid = (node[x].l + node[x].r) / 2;    if (k <= mid) set(k, v, lson(x));    if (k > mid) set(k, v, rson(x));    node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min);}int query(int l, int r, int x = 0) {    if (node[x].l >= l && node[x].r <= r)return node[x].Min;    int mid = (node[x].l + node[x].r) / 2;    int ans = INF;    if (l <= mid) ans = min(ans, query(l, r, lson(x)));    if (r > mid) ans = min(ans, query(l, r, rson(x)));    return ans;}int main() {    while (~scanf("%d%d", &n, &m)) {for (int i = 1; i <= n; i++)    scanf("%d", &num[i]);build(1, n);while (m--) {    scanf("%s", Q);    handle(Q);    if (Q[0] == 'q')printf("%d\n", query(shif[0], shif[1]));    else {int tmp = num[shif[0]];set(shif[sn - 1], num[shif[0]]);for (int i = 1; i < sn; i++) {    set(shif[i - 1], num[shif[i]]);    num[shif[i - 1]] = num[shif[i]];}num[shif[sn - 1]] = tmp;    }}    }    return 0;}