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  • 「NOI2005」维护数列

    「NOI2005」维护数列

    传送门
    维护过程有点像线段树。
    但我们知道线段树的节点并不是实际节点,而平衡树的节点是实际节点。
    所以在向上合并信息时要加入根节点信息。
    然后节点再删除后编号要回退(栈),不然会爆空间。
    具体实现看代码就好了。
    参考代码:

    #include <algorithm>
    #include <cstdlib>
    #include <cstdio>
    #define rg register
    #define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
    using namespace std;
    template < class T > inline void read(T& s) {
    	s = 0; int f = 0; char c = getchar();
    	while ('0' > c || c > '9') f |= c == '-', c = getchar();
    	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    	s = f ? -s : s;
    }
    
    const int _ = 1e6 + 2;
    
    int tot, tmp[_];
    int n, q, rt, siz[_], pri[_], ch[2][_];
    int val[_], mx[_], L[_], R[_], sum[_], rev[_], tag[_], flag[_];
    
    inline int Newnode(int v) {
    	int id = tmp[tot--];
    	siz[id] = 1, pri[id] = rand();
    	val[id] = mx[id] = sum[id] = v, L[id] = R[id] = max(0, v);
    	ch[0][id] = ch[1][id] = rev[id] = flag[id] = 0;
    	return id;
    }
    
    inline void pushup(int p) {
    	siz[p] = siz[ch[0][p]] + siz[ch[1][p]] + 1;
    	sum[p] = sum[ch[0][p]] + sum[ch[1][p]] + val[p];
    	L[p] = max(0, max(L[ch[0][p]], sum[ch[0][p]] + val[p] + L[ch[1][p]]));
    	R[p] = max(0, max(R[ch[1][p]], sum[ch[1][p]] + val[p] + R[ch[0][p]]));
    	mx[p] = max(val[p], R[ch[0][p]] + val[p] + L[ch[1][p]]);
    	if (ch[0][p]) mx[p] = max(mx[p], mx[ch[0][p]]);
    	if (ch[1][p]) mx[p] = max(mx[p], mx[ch[1][p]]);
    }
    
    inline void Rev(int p) {
    	rev[p] ^= 1, swap(L[p], R[p]), swap(ch[0][p], ch[1][p]);
    }
    
    inline void Tag(int p, int v) {
    	flag[p] = 1, val[p] = tag[p] = v, sum[p] = siz[p] * v;
    	L[p] = R[p] = max(0, sum[p]), mx[p] = max(val[p], sum[p]);
    }
    
    inline void pushdown(int p) {
    	if (rev[p]) {
    		if (ch[0][p]) Rev(ch[0][p]);
    		if (ch[1][p]) Rev(ch[1][p]);
    		rev[p] = 0;
    	}
    	if (flag[p]) {
    		if (ch[0][p]) Tag(ch[0][p], tag[p]);
    		if (ch[1][p]) Tag(ch[1][p], tag[p]);
    		flag[p] = 0;
    	}
    }
    
    inline int merge(int x, int y) {
    	if (!x || !y) return x + y;
    	if (pri[x] > pri[y])
    		return pushdown(x), ch[1][x] = merge(ch[1][x], y), pushup(x), x;
    	else
    		return pushdown(y), ch[0][y] = merge(x, ch[0][y]), pushup(y), y;
    }
    
    inline void split(int p, int k, int& x, int& y) {
    	if (p) pushdown(p);
    	if (!p) { x = y = 0; return ; }
    	if (siz[ch[0][p]] + 1 <= k)
    		return x = p, split(ch[1][p], k - siz[ch[0][p]] - 1, ch[1][x], y), pushup(p);
    	else
    		return y = p, split(ch[0][p], k, x, ch[0][y]), pushup(p);
    }
    
    inline void erase(int p) {
    	if (!p) return ;
    	tmp[++tot] = p;
    	if (ch[0][p]) erase(ch[0][p]);
    	if (ch[1][p]) erase(ch[1][p]);
    }
    
    int main() {
    	srand((unsigned long long) new char);
    	read(n), read(q);
    	for (rg int i = 1; i <= 500000; ++i) tmp[++tot] = i;
    	for (rg int v, i = 1; i <= n; ++i) read(v), rt = merge(rt, Newnode(v));
    	char s[15];
    	for (int pos, x, v, a, b, c; q--; ) {
    		scanf("%s", s);
    		if (s[0] == 'I') {
    			read(pos), read(x);
    			split(rt, pos, a, b);
    			while (x--) read(c), a = merge(a, Newnode(c));
    			rt = merge(a, b);
    		}
    		if (s[0] == 'D') {
    			read(pos), read(x);
    			split(rt, pos - 1, a, b);
    			split(b, x, b, c);
    			erase(b);
    			rt = merge(a, c);
    		}
    		if (s[0] == 'M' && s[2] == 'K') {
    			read(pos), read(x), read(v);
    			split(rt, pos - 1, a, b);
    			split(b, x, b, c);
    			Tag(b, v);
    			rt = merge(a, merge(b, c));
    		}
    		if (s[0] == 'R') {
    			read(pos), read(x);
    			split(rt, pos - 1, a, b);
    			split(b, x, b, c);
    			Rev(b);
    			rt = merge(a, merge(b, c));
    		}
    		if (s[0] == 'G') {
    			read(pos), read(x);
    			split(rt, pos - 1, a, b);
    			split(b, x, b, c);
    			printf("%d
    ", sum[b]);
    			rt = merge(a, merge(b, c));
    		}
    		if (s[0] == 'M' && s[2] == 'X')
    			printf("%d
    ", mx[rt]);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zsbzsb/p/12231597.html
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