G - Periodic RMQ Problem
思路：
题目给一段序列，然后序列复制很多次；
维护序列很多次后的性质；
线段树动态开点；
来，上代码：
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm>using namespace std;#define maxn 100005struct TreeNodeType {int l, r, mid, min, flag;TreeNodeType *lc, *rc;TreeNodeType(){flag=0;lc = NULL;rc = NULL;} }; struct TreeNodeType *root, *rot;int n, k, m;inline void in(int &now) {char Cget = getchar(); now = 0;while (Cget > '9' || Cget < '0') Cget = getchar();while (Cget >= '0'&&Cget <= '9'){now = now * 10 + Cget - '0';Cget = getchar();} }void tree_build_ori(TreeNodeType *&now, int l, int r) {if (now == NULL){now = new TreeNodeType;now->l = l, now->r = r;now->mid = (l + r) >> 1;}if (l == r){in(now->min);return;}tree_build_ori(now->lc, l, now->mid);tree_build_ori(now->rc, now->mid + 1, r);now->min = min(now->lc->min, now->rc->min); }int tree_query_ori(TreeNodeType *&now, int l, int r) {if (now->l == l&&now->r == r) return now->min;if (l > now->mid) return tree_query_ori(now->rc, l, r);else if (r <= now->mid) return tree_query_ori(now->lc, l, r);else return min(tree_query_ori(now->lc, l, now->mid), tree_query_ori(now->rc, now->mid + 1, r)); }inline void tree_down(TreeNodeType *&now) {now->lc->min = now->flag;now->lc->flag = now->flag;now->rc->min = now->flag;now->rc->flag = now->flag;now->flag = 0; }int solve(int l, int r) {if(r-l+1>=n) return rot->min;l%=n,r%=n;if(l==0) l=n;if(r==0) r=n;if(r<l) return min(tree_query_ori(rot, l, n), tree_query_ori(rot, 1, r));else return tree_query_ori(rot,l,r); }void tree_change(TreeNodeType *&now, int l, int r, int x) {if (now->l == l&&now->r == r){now->min = x;now->flag = x;return;}if (now->rc == NULL){now->rc = new TreeNodeType;now->rc->l = now->mid + 1;now->rc->r = now->r;now->rc->mid = (now->rc->r + now->rc->l) >> 1;now->rc->min = solve(now->rc->l, now->rc->r);}if (now->lc == NULL){now->lc = new TreeNodeType;now->lc->l = now->l;now->lc->r = now->mid;now->lc->mid = (now->lc->l + now->lc->r) >> 1;now->lc->min = solve(now->lc->l, now->lc->r);}if (now->flag) tree_down(now);if (l > now->mid) tree_change(now->rc, l, r, x);else if (r <= now->mid) tree_change(now->lc, l, r, x);else{tree_change(now->lc, l, now->mid, x);tree_change(now->rc, now->mid + 1, r, x);}now->min = min(now->lc->min, now->rc->min); }int tree_query(TreeNodeType *&now, int l, int r) {if (now->l == l&&now->r == r) return now->min;if (now->rc == NULL){now->rc = new TreeNodeType;now->rc->l = now->mid + 1;now->rc->r = now->r;now->rc->mid = (now->rc->r + now->rc->l) >> 1;now->rc->min = solve(now->rc->l, now->rc->r);}if (now->lc == NULL){now->lc = new TreeNodeType;now->lc->l = now->l;now->lc->r = now->mid;now->lc->mid = (now->lc->l + now->lc->r) >> 1;now->lc->min = solve(now->lc->l, now->lc->r);}if (now->flag) tree_down(now);if (l > now->mid) return tree_query(now->rc, l, r);else if (r <= now->mid) return tree_query(now->lc, l, r);else return min(tree_query(now->lc, l, now->mid), tree_query(now->rc, now->mid + 1, r));now->min = min(now->lc->min, now->rc->min); }int main() {root = NULL, rot = NULL; int op, l, r, x;in(n), in(k), tree_build_ori(rot, 1, n), in(m);root = new TreeNodeType;root->l = 1, root->r = n*k, root->mid = 1 + n*k >> 1, root->min = rot->min;for (; m--;){in(op), in(l), in(r);if (op == 2) printf("%d\n", tree_query(root, l, r));else in(x),tree_change(root, l, r, x);}return 0; }
