[洛谷P4940]Portal2

题目大意：维护两个栈，几个操作：

1. $PUSH\;x\;num：$把$num$压入栈$x$
2. $POP\;x：$弹出栈$x$栈顶元素
3. $ADD\;x：$取出两个栈栈顶，把相加值压入栈$x$
4. $SUB\;x：$取出两个栈栈顶，把相减的绝对值压入栈$x$
5. $DEL\;x：$清空栈$x$
6. $MOVE\;x\;y：$把$y$中的元素一个个弹出，依次压入栈$x$
7. $SWAP：$交换两个栈
8. $END：$结束

题解：可以平衡树搞，但是也可以用双向链表。两个双向链表，除了第$6$个，其他都是基本操作。第$6$个操作时，就把$y$的栈顶与$x$的栈顶接起来，把$x$的栈顶设为$y$的栈底就行了

卡点：莫名$RE$了很多次（包括后面$WA$的），开大数组后通过（我也不知道为什么，中间尝试查错，没有成功）

C++ Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#define maxn 7000010
#define TANG_Yx 20040826
struct node {
	long long w;
	int nxt[2];
	inline int find(int pos = TANG_Yx) {
		if (pos == TANG_Yx) {for (int i = 0; i < 2; i++) if (nxt[i] != 0) return nxt[i];}
		else for (int i = 0; i < 2; i++) if (nxt[i] == pos) return nxt[i];
		return TANG_Yx;
	}
	inline int find_pos(int pos = TANG_Yx) {
		if (pos == TANG_Yx) {for (int i = 0; i < 2; i++) if (nxt[i] != 0) return i;}
		else for (int i = 0; i < 2; i++) if (nxt[i] == pos) return i;
		return TANG_Yx;
	}
	inline void clear_nxt(int pos) {
		int __pos = find_pos(pos);
		if (__pos != TANG_Yx) nxt[__pos] = 0;
	}
};

node S[maxn << 1];
struct Stack {
	int begin, end, idx, sz = 0;
	inline Stack(int __idx = 0) {
		begin = end = 0;
		idx = __idx;
		sz = 0;
	}
	inline bool empty() {return sz < 1;}
	inline void clear() {
		begin = end = 0;
		sz = 0;
	}
	inline long long top() {
		if (empty()) return TANG_Yx;
		return S[begin].w;
	}
	inline int pop() {
		if (empty()) return TANG_Yx;
		int __begin = S[begin].find();
		sz--;
		if (sz) {
			S[__begin].clear_nxt(begin);
			begin = __begin;
		} else begin = 0;
		return sz;
	}
	inline void push(long long num) {
		int __begin = ++idx;
		S[__begin].w = num;
		S[__begin].nxt[0] = begin;
		S[__begin].nxt[1] = 0;
		int __nxt = S[begin].find_pos(0);
		S[begin].nxt[__nxt] = __begin;
		begin = __begin;
		if (!sz) end = __begin;
		sz++;
	}
} sta[2];
inline long long abs(long long a) {return a < 0 ? -a : a;}

int opt = 0, result;
int main() {
	sta[0] = Stack();
	sta[1] = Stack(maxn);
	while (true) {
		char op[10];
		int x; long long y;
		scanf("%s", op);
		if (strcmp(op, "PUSH") == 0) {
			scanf("%d%lld", &x, &y);
			sta[x ^ opt].push(y);
		}
		if (strcmp(op, "POP") == 0) {
			scanf("%d", &x);
			result = sta[x ^ opt].pop();
			if (result == TANG_Yx) printf("UN");
		}
		if (strcmp(op, "ADD") == 0) {
			scanf("%d", &x);
			long long l = sta[0].top(), r = sta[1].top();
			if (l != TANG_Yx && r != TANG_Yx) {
				sta[0].pop(), sta[1].pop();
				sta[x ^ opt].push(l + r);
			} else printf("UN");
		}
		if (strcmp(op, "SUB") == 0) {
			scanf("%d", &x);
			long long l = sta[0].top(), r = sta[1].top();
			if (l != TANG_Yx && r != TANG_Yx) {
				sta[0].pop(), sta[1].pop();
				sta[x ^ opt].push(abs(l - r));
			} else printf("UN");
		}
		if (strcmp(op, "DEL") == 0) {
			scanf("%d", &x);
			sta[x ^ opt].clear();
		}
		if (strcmp(op, "MOVE") == 0) {
			int l, r;
			scanf("%d%d", &l, &r);
			l ^= opt, r ^= opt;
			if (!sta[r].empty()) {
				if (sta[l].empty()) {
					sta[l].begin = sta[r].end;
					sta[l].end = sta[r].begin;
					sta[l].sz = sta[r].sz;
				} else {
					int tmp = S[sta[l].begin].find_pos(0);
					S[sta[l].begin].nxt[tmp] = sta[r].begin;
					tmp = S[sta[r].begin].find_pos(0);
					S[sta[r].begin].nxt[tmp] = sta[l].begin;
					sta[l].begin = sta[r].end;
					sta[l].sz += sta[r].sz;
				}
				sta[r].clear();
			}
		}
		if (strcmp(op, "SWAP") == 0) {
			opt ^= 1;
		}
		puts("SUCCESS");
		if (strcmp(op, "END") == 0) {
			break;
		}
	}
	bool flag = false;
	while (sta[opt].top() != TANG_Yx) {
		flag = true;
		printf("%lld ", sta[opt].top());
		sta[opt].pop();
	}
	if (!flag) printf("NONE");
	puts("");
	flag = false;
	while (sta[opt ^ 1].top() != TANG_Yx) {
		flag = true;
		printf("%lld ", sta[opt ^ 1].top());
		sta[opt ^ 1].pop();
	}
	if (!flag) printf("NONE");
	puts("");
	return 0;
}