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  • bzoj2959: 长跑(LCT+并查集)

    题解

    动态树Link-cut tree(LCT)总结

    LCT常数大得真实

    没有环,就是(lct)裸题吧

    有环,我们就可以绕环转一圈,缩点

    怎么搞?

    当形成环时,把所有点的值全部加到一个点上,用并查集维护加到哪个点上

    判断连通性再用一个并查集

    Code

    #include<bits/stdc++.h>
    
    #define LL long long
    #define RG register
    
    using namespace std;
    template<class T> inline void read(T &x) {
    	x = 0; RG char c = getchar(); bool f = 0;
    	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    	x = f ? -x : x;
    	return ;
    }
    template<class T> inline void write(T x) {
    	if (!x) {putchar(48);return ;}
    	if (x < 0) x = -x, putchar('-');
    	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
    }
    const int N = 150010;
    struct node {
    	int v, s, fa, ch[2];
    	bool rev;
    }t[N];
    int S[N], top;
    #define lson (t[x].ch[0])
    #define rson (t[x].ch[1])
    void pushup(int x) {t[x].s = (t[lson].s + t[rson].s + t[x].v);return ;}
    void putrev(int x) {
    	swap(lson, rson);
    	t[x].rev ^= 1;
    	return ;
    }
    void pushdown(int x) {
    	if (t[x].rev) {
    		if (lson) putrev(lson);
    		if (rson) putrev(rson);
    		t[x].rev = 0;
    	}
    }
    int fa[N], Fa[N];
    int find(int x) {
    	return fa[x] == x ? x : fa[x] = find(fa[x]);
    }
    int Find(int x) {
    	return Fa[x] == x ? x : Fa[x] = Find(Fa[x]);
    }
    bool isroot(int x) {
    	return (t[find(t[x].fa)].ch[0] != x) && (t[find(t[x].fa)].ch[1] != x);
    }
    
    void rotate(int x) {
    	int y = find(t[x].fa), z = find(t[y].fa), k = t[y].ch[1] == x;
    	if (!isroot(y)) t[z].ch[t[z].ch[1] == y] = x;
    	t[x].fa = z;
    	t[t[x].ch[k^1]].fa = y, t[y].ch[k] = t[x].ch[k^1];
    	t[x].ch[k^1] = y; t[y].fa = x; 
    	pushup(y);
    	return ;
    }
    
    void splay(int x) {
    	S[top = 1] = x;
    	for (RG int i = x; !isroot(i); i = find(t[i].fa)) S[++top] = find(t[i].fa);
    	for (RG int i = top; i; i--) pushdown(S[i]);
    	while (!isroot(x)) {
    		int y = find(t[x].fa), z = find(t[y].fa);
    		if (!isroot(y))
    			(t[y].ch[1] == x) ^ (t[z].ch[1] == y) ? rotate(x) : rotate(y);
    		rotate(x);
    	}
    	pushup(x);
    	return ;
    }
    
    void access(int x) {
    	for (int y = 0; x; y = x, x = find(t[x].fa))
    		splay(x), t[x].ch[1] = y, pushup(x); 
    }
    
    void makeroot(int x) {
    	access(x); splay(x); putrev(x);
    	return ;
    }
    
    void link(int x, int y) {
    	makeroot(x);
    	t[x].fa = y;
    }
    
    void split(int x, int y) {
    	makeroot(x), access(y), splay(y);
    	return ;
    }
    int a[N], n, m;
    
    void dfs(int x, int y) {
    	fa[x] = y;
    	pushdown(x); 
    	if (lson) dfs(lson, y);
    	if (rson) dfs(rson, y);
    	return ;
    }
    
    int main() {
    	read(n), read(m);
    	for (int i = 1; i <= n; i++) read(a[i]), fa[i] = Fa[i] = i, t[i].s = t[i].v = a[i];
    	while (m--) {
    		int x, y, op;
    		read(op), read(x), read(y);
    		if (op == 1) {
    			x = find(x), y = find(y);
    			if (x == y) continue;
    			if (Find(x) != Find(y)) {
    				link(x, y);
    				Fa[Find(y)] = Find(x);
    			}
    			else {
    				split(x, y);
    				t[y].v = t[y].s;
    				dfs(y, y);
    				t[y].ch[0] = 0;
    			}
    		}
    		else if (op == 2) {
    			int tx = find(x); 
    			splay(tx);
    			t[tx].s += y - a[x];
    			t[tx].v += y - a[x];
    			a[x] = y;
    		}
    		else {
    			x = find(x), y = find(y);
    			if (Find(x) != Find(y)) {
    				puts("-1");
    				continue;
    			}
    			split(x, y);
    			printf("%d
    ", t[y].s);
    		}
    	}	
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/zzy2005/p/10324298.html
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