题意
给定$n$个数，$m$个询问，每次在$[L,R]$区间加上一个数，或者反转一个区间$[L,R]$，或者循环右移区间$[L,R]$共$T$次，或者在第$x$个数后插入一个数$p$，或者删除第$x$个数，或者查询区间$L,R$的最小值
Splay模板题
对于区间操作，利用splay提取区间后打懒标记更新，单点操作看作是对$[L,L+1]$区间的操作
时间复杂度$O(m \log n)$
代码
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define inf 0x7f7f7f7f
using namespace std;
const int N = 800005;
int n, m, a[N];
int ch[N][2], fa[N], key[N], rev[N], sz[N], Min[N], add[N];
int root, tot;
inline int get(int x) {return ch[fa[x]][1] == x;} // left is 0, right is 1
inline void pushup(int x) {
sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1;
Min[x] = key[x];
if(ch[x][0]) Min[x] = min(Min[x], Min[ch[x][0]]);
if(ch[x][1]) Min[x] = min(Min[x], Min[ch[x][1]]);
}
inline void update_add(int x,int add_) {key[x] += add_; add[x] += add_; Min[x] += add_;}
inline void pushdown(int x) {
if(!x) return;
if(add[x]) {
if(ch[x][0]) {update_add(ch[x][0], add[x]);}
if(ch[x][1]) {update_add(ch[x][1], add[x]);}
add[x] = 0;
}
if(rev[x]) {
rev[x] = 0;
swap(ch[x][0], ch[x][1]);
rev[ch[x][0]] ^= 1; rev[ch[x][1]] ^= 1;
}
}
inline void rotate(int x) {
int f = fa[x], ff = fa[f], which = get(x);
pushdown(f); pushdown(x);
ch[f][which] = ch[x][which ^ 1];
fa[ch[f][which]] = f;
ch[x][which ^ 1] = f;
fa[f] = x; fa[x] = ff;
if(ff) ch[ff][ch[ff][1] == f] = x;
pushup(f); pushup(x);
}
inline void splay(int x, int target) {
while(fa[x] != target) {
if(fa[fa[x]] != target) {
rotate((get(x) == get(fa[x])) ? fa[x] : x);
}
rotate(x);
}
if(!target) root = x;
}
inline int newnode(int v, int f) {
int x = ++tot;
ch[x][0] = ch[x][1] = 0; fa[x] = f;
key[x] = Min[x] = v; sz[x] = 1; add[x] = rev[x] = 0;
return x;
}
int build(int l, int r, int f) {
if(l > r) return 0;
int mid = (l + r) / 2;
int x = newnode(a[mid], f);
ch[x][0] = build(l, mid - 1, x);
ch[x][1] = build(mid + 1, r, x);
pushup(x);
return x;
}
inline void init() {tot = 0; root = build(0, n + 1, 0);}
inline int get_kth(int x, int k) {
if(!x) return 0;
while(1) {
pushdown(x);
if(k == sz[ch[x][0]] + 1) break;
if(k > sz[ch[x][0]] + 1) {
k -= sz[ch[x][0]] + 1; x = ch[x][1];
}else x = ch[x][0];
}
return x;
}
char opt[20]; int L, R, d;
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
a[0] = -inf; a[n + 1] = inf;
init();
scanf("%d", &m);
for(int i = 1; i <= m; ++i) {
scanf("%s", opt);
if(!strcmp(opt, "ADD")) {
scanf("%d%d%d", &L, &R, &d);
splay(get_kth(root, L), 0); splay(get_kth(root, R + 2), root);
update_add(ch[ch[root][1]][0], d);
pushup(ch[root][1]); pushup(root);
}else if(!strcmp(opt, "REVERSE")) {
scanf("%d%d", &L, &R);
splay(get_kth(root, L), 0); splay(get_kth(root, R + 2), root);
rev[ch[ch[root][1]][0]] ^= 1;
pushup(ch[root][1]); pushup(root);
}else if(!strcmp(opt, "REVOLVE")) {
scanf("%d%d%d", &L, &R, &d);
d = (d % (R - L + 1) + (R - L + 1)) % (R - L + 1);
if(!d) continue;
splay(get_kth(root, R - d + 1), 0); splay(get_kth(root, R + 2), root);
int tmp = ch[ch[root][1]][0];
ch[ch[root][1]][0] = 0;
pushup(ch[root][1]); pushup(root);
splay(get_kth(root, L), 0); splay(get_kth(root, L + 1), root);
ch[ch[root][1]][0] = tmp; fa[tmp] = ch[root][1];
pushup(ch[root][1]); pushup(root);
}else if(!strcmp(opt, "INSERT")) {
scanf("%d%d", &L, &R);
splay(get_kth(root, L + 1), 0); splay(get_kth(root, L + 2), root);
ch[ch[root][1]][0] = newnode(R, ch[root][1]);
pushup(ch[root][1]); pushup(root);
}else if(!strcmp(opt, "DELETE")) {
scanf("%d%d", &L);
splay(get_kth(root, L), 0); splay(get_kth(root, L + 2), root);
ch[ch[root][1]][0] = 0;
pushup(ch[root][1]); pushup(root);
}else if(!strcmp(opt, "MIN")) {
scanf("%d%d", &L, &R);
splay(get_kth(root, L), 0); splay(get_kth(root, R + 2), root);
printf("%d\n", Min[ch[ch[root][1]][0]]);
}
}
return 0;
} 转载于:https://www.cnblogs.com/ogiso-setsuna/p/7989510.html