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  • 【洛谷 P4219】 [BJOI2014]大融合(LCT)

    题目链接
    维护子树信息向来不是(LCT)所擅长的,所以我没搞懂qwq
    权当背背模板吧。Flash巨佬的blog里面写了虽然我没看懂

    #include <cstdio>
    #define R register int
    #define I inline void
    #define lc c[x][0]
    #define rc c[x][1]
    const int MAXN = 300010;
    inline int read(){
        int s = 0, w = 1;
        char ch = getchar();
        while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
        while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
        return s * w;
    }
    int f[MAXN], c[MAXN][2], v[MAXN], sz[MAXN], st[MAXN], tag[MAXN], ss[MAXN];
    inline int nroot(R x){
    	return c[f[x]][0] == x || c[f[x]][1] == x;
    }
    I pushup(R x){
    	sz[x] = sz[lc] + sz[rc] + ss[x] + 1;
    }
    I pushdown(R x){
    	if(tag[x]){
    		R t = lc; lc = rc; rc = t;
    		tag[lc] ^= 1; tag[rc] ^= 1; tag[x] = 0;
    	}
    }
    I rotate(R x){
    	R y = f[x], z = f[y], k = c[y][1] == x, w = c[x][!k];
    	if(nroot(y)) c[z][c[z][1] == y] = x;
    	c[x][!k] = y; c[y][k] = w; f[y] = x; f[x] = z;
    	if(w) f[w] = y;
    	pushup(y);
    }
    I pushall(R x){
        if(nroot(x)) pushall(f[x]);
        pushdown(x);
    }
    I splay(R x){
    	pushall(x);
    	while(nroot(x)) rotate(x);
    	pushup(x);
    }
    I access(R x){
    	for(R y = 0; x; x = f[y = x]){
    	   splay(x); ss[x] += sz[rc]; ss[x] -= sz[rc = y]; pushup(x);
        }
    }
    I makeroot(R x){
    	access(x); splay(x); 
    	tag[x] ^= 1;
    }
    I split(R x, R y){
    	makeroot(x); access(y); splay(y);
    }
    I link(R x, R y){
    	split(x, y);
    	ss[f[x] = y] += sz[x];
    	pushup(y);
    }
    int n, m, a, b;
    char opt;
    int main(){
    	n = read(); m = read();
    	for(R i = 1; i <= n; ++i) sz[i] = 1;
    	while(m--){
    		opt = getchar(); while(opt != 'A' && opt != 'Q') opt = getchar();
    		 a = read(); b = read();
    		switch(opt){
    			case 'Q' : split(a, b); printf("%lld
    ", (long long)(ss[a] + 1) * (ss[b] + 1)); break;
    			case 'A' : link(a, b); break;
    		}
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Qihoo360/p/10360978.html
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