SPOJ GSS6 Can you answer these queries VI

题意：

要求维护一个数列， 满足插入， 删除，修改和询问最大连续子段和的操作。。

思路：

用splay维护， 由于子段要求非空。。 所以push_up那里不能直接用空儿子的值。。

#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <cmath>#include <queue>#include <iomanip>#include <map>#include <set>#include <algorithm>using namespace std;#define N 200020#define mod 1000000007#define mxe 2000020#define MP make_pair#define PB push_back#define LL long long#define ULL unsigned LL#define inf 0x3f3f3f3f#define lson ll, md, ls#define rson md + 1, rr, rs#pragma comment(linker, "/STACK:1024000000,1024000000")int pre[N], ch[N][2], key[N], sz[N];int mx[N], lx[N], rx[N], sum[N];int tot, root;int n, a[N], m;int creat(int p, int val) {int x = ++tot;ch[x][0] = ch[x][1] = 0;pre[x] = p;key[x] = val;sz[x] = 1;mx[x] = lx[x] = rx[x] = sum[x] = val;return x;}void push_up(int x) {int lc = ch[x][0], rc = ch[x][1];sum[x] = sum[lc] + sum[rc] + key[x];sz[x] = sz[lc] + sz[rc] + 1;if(lc == 0 && rc == 0) {mx[x] = lx[x] = rx[x] = key[x];return;}if(rc == 0) {mx[x] = max(key[x], mx[lc]);mx[x] = max(mx[x], key[x] + rx[lc]);lx[x] = max(lx[lc], sum[lc] + key[x]);rx[x] = max(key[x], key[x] + rx[lc]);return ;}if(lc == 0) {mx[x] = max(key[x], mx[rc]);mx[x] = max(mx[x], key[x] + lx[rc]);lx[x] = max(key[x], key[x] + lx[rc]);rx[x] = max(rx[rc], sum[rc] + key[x]);return;}mx[x] = max(mx[lc], mx[rc]);mx[x] = max(mx[x], key[x]);mx[x] = max(mx[x], key[x] + rx[lc]);mx[x] = max(mx[x], key[x] + lx[rc]);mx[x] = max(mx[x], key[x] + rx[lc] + lx[rc]);lx[x] = max(lx[lc], sum[lc] + key[x]);lx[x] = max(lx[x], sum[lc] + key[x] + lx[rc]);rx[x] = max(rx[rc], sum[rc] + key[x]);rx[x] = max(rx[x], sum[rc] + key[x] + rx[lc]);}int build(int l, int r, int p) {if(l > r) return 0;int mm = (l + r) / 2;int x = creat(p, a[mm]);ch[x][0] = build(l, mm - 1, x);ch[x][1] = build(mm + 1, r, x);push_up(x);return x;}void rot(int x) {int y = pre[x], d = ch[y][1] == x;ch[y][d] = ch[x][!d];if(ch[x][!d]) pre[ch[x][!d]] = y;ch[x][!d] = y;pre[x] = pre[y];pre[y] = x;if(pre[x]) ch[pre[x]][ch[pre[x]][1]==y] = x;push_up(y);}void splay(int x, int goal) {while(pre[x] != goal) {int f = pre[x], ff = pre[f];if(ff == goal)rot(x);else if((ch[ff][1] == f) == (ch[f][1] == x))rot(f), rot(x);elserot(x), rot(x);}push_up(x);if(goal == 0) root = x;}int kth(int k) {int x = root;while(1) {if(k <= sz[ch[x][0]]) x = ch[x][0];else {k -= sz[ch[x][0]] + 1;if(k == 0) return x;x = ch[x][1];}}return -1;}int main() {//freopen("tt.txt", "r", stdin);while(scanf("%d", &n) != EOF) {tot = 0;for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);root = build(0, n + 1, 0);scanf("%d", &m);while(m--) {char op[5];int u, v, x, y;scanf("%s%d", op, &u);if(op[0] != 'D') scanf("%d", &v);if(op[0] == 'I') {splay(kth(u), 0);splay(kth(u + 1), root);int y = ch[root][1];int x = creat(y, v);ch[y][0] = x;push_up(y);push_up(root);}if(op[0] == 'D') {splay(kth(u), 0);splay(kth(u + 1), root);int y = ch[root][1];ch[root][1] = ch[y][1];pre[ch[y][1]] = root;push_up(root);}if(op[0] == 'R') {splay(kth(u + 1), 0);key[root] = v;push_up(root);}if(op[0] == 'Q') {splay(kth(u), 0);splay(kth(v + 2), root);y = ch[root][1];x = ch[y][0];printf("%d\n", mx[x]);}}}return 0;}