• BZOJ2243 [SDOI2011]染色(树链剖分+线段树合并)


    题目链接 BZOJ2243

    树链剖分 $+$ 线段树

    线段树每个节点维护$lc$, $rc$, $s$

    $lc$代表该区间的最左端的颜色,$rc$代表该区间的最右端的颜色

    $s$代表该区间的所有连续颜色段数(仅考虑该区间)

    $lazy$表示延迟信息。

    #include <bits/stdc++.h>
    
    using namespace std;
    
    #define rep(i, a, b)	for (int i(a); i <= (b); ++i)
    #define dec(i, a, b)	for (int i(a); i >= (b); --i)
    #define lson		i << 1, L, mid
    #define rson		i << 1 | 1, mid + 1, R
    
    typedef long long LL;
    
    const int N = 1e5 + 10;
    const int A = 21;
    
    vector <int> v[N];
    int n, m;
    int c[N];
    int f[N][A];
    int lazy[N << 2], s[N << 2], lc[N << 2], rc[N << 2];
    int h[N], deep[N], sz[N];
    int son[N];
    int tot = 0;
    int top[N];
    
    
    int LCA(int a, int b){
    	if (deep[a] < deep[b]) swap(a, b);
    	for (int i = 0,  delta = deep[a] - deep[b]; delta; delta >>= 1, ++i) if (delta & 1) a = f[a][i];
    	if (a == b) return a;
    	dec(i, 19, 0) if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];
    	return f[a][0];
    }
    
    inline void pushup(int i){
    	lc[i] = lc[i << 1];
    	rc[i] = rc[i << 1 | 1];
    	if (rc[i << 1] ^ lc[i << 1 | 1]) s[i] = s[i << 1] + s[i << 1 | 1];
    	else s[i] = s[i << 1] + s[i << 1 | 1] - 1;
    }
    
    inline void pushdown(int i, int L, int R){
    	int tmp = lazy[i];
    	if (tmp == -1 || L == R) return;
    
    	s[i << 1]      = s[i << 1 | 1]    = 1;
    	lazy[i << 1]   = lazy[i << 1 | 1] = tmp;
    
    	lc[i << 1]     = rc[i << 1]       = tmp;
    	lc[i << 1 | 1] = rc[i << 1 | 1]   = tmp;
    	lazy[i] = -1;
    }	
    
    void dfs1(int x, int fa, int dep){
    	sz[x] = 1;
    	deep[x] = dep;
    
    	if (fa){ 
                    f[x][0] = fa;
    		for (int i = 0; f[f[x][i]][i]; ++i) f[x][i + 1] = f[f[x][i]][i];
    	}
    
    	int ct = (int)v[x].size();
    
    	rep(i, 0, ct - 1){
    		int u = v[x][i];
    		if (u == fa) continue;
    		dfs1(u, x, dep + 1);
    		sz[x] += sz[u];
    		if (sz[son[x]] < sz[u]) son[x] = u;
    	}
    }
    
    void dfs2(int x, int tp){
    	h[x] = ++tot;
    	top[x] = tp;
    	if (son[x]) dfs2(son[x], tp);
    	
    	int ct = (int)v[x].size();
    	rep(i, 0, ct - 1){
    		int u = v[x][i];
    		if (u == f[x][0] || u == son[x]) continue;
    		dfs2(u, u);
    	}
    }
    
    void build(int i, int L, int R){
    	s[i] = 1;
    	lazy[i] = -1;
    
    	if (L == R) return;
    	int mid = (L + R) >> 1;
    
    	build(lson);
    	build(rson);
    }
    
    void change(int i, int L, int R, int l, int r, int val){
    	pushdown(i, L, R);
    	if (L == l && R == r){
    		lc[i] = rc[i] = val;
    		s[i]  = 1;
    		lazy[i] = val;
    		return;
    	}
    
    	int mid = (L + R) >> 1;
    	if (r <= mid) change(lson, l, r, val);
    	else if (l > mid) change(rson, l, r, val);
    	else{
    		change(lson, l, mid, val);
    		change(rson, mid + 1, r, val);
    	}
    
    	pushup(i);
    }
    
    int query(int i, int L, int R, int l, int r){
    	pushdown(i, L, R);
    	if (L == l && R == r) return s[i];
    
    	int mid = (L + R) >> 1;
    	if (r <= mid) return query(lson, l, r);
    	else if (l > mid) return query(rson, l, r);
    	else{
    		int tmp = 1;
    		if (rc[i << 1] ^ lc[i << 1 | 1]) tmp = 0;
    		return query(lson, l, mid) + query(rson, mid + 1, r) - tmp;
    	}
    }
    
    int getcolor(int i, int L, int R, int x){
    	pushdown(i, L, R);
    	if (L == R) return lc[i];
    	int mid = (L + R) >> 1;
    	if (x <= mid) return getcolor(lson, x);
    	else return getcolor(rson, x);
    }
    
    int solvesum(int x, int tp){
    	int ret = 0;
    	for (; top[x] ^ top[tp] ;){
    		ret += query(1, 1, n, h[top[x]], h[x]);
    		if (getcolor(1, 1, n, h[top[x]]) == getcolor(1, 1, n, h[f[top[x]][0]])) --ret;
    		x = f[top[x]][0];
    	}
    
    	ret += query(1, 1, n, h[tp], h[x]);
    	return ret;
    }
    
    void solvechange(int x, int tp, int val){
    	for (; top[x] ^ top[tp]; ){
    		change(1, 1, n, h[top[x]], h[x], val);
    		x = f[top[x]][0];
    	}
    
    	change(1, 1, n, h[tp], h[x], val);
    }
    
    
    void solve(){
    	int x, y, z;
    	dfs1(1, 0, 0);
    	dfs2(1, 1);
    	build(1, 1, n);
    	rep(i, 1, n) change(1, 1, n, h[i], h[i], c[i]);
    
    	rep(i, 1, m){
    		char ch[10];
    		scanf("%s", ch);
    		if (ch[0] == 'Q'){
    			scanf("%d%d", &x, &y);
    			int t = LCA(x, y);
    			printf("%d
    ", solvesum(x, t) + solvesum(y, t) - 1);
    		}
    		else{
    			scanf("%d%d%d", &x, &y, &z);
    			int t = LCA(x, y);
    			solvechange(x, t, z);
    			solvechange(y, t, z);
    		}
    	}
    }
    
    void init(){
    	scanf("%d%d", &n, &m);
    	rep(i, 1, n) scanf("%d", c + i);
    	rep(i, 2, n){
    		int x, y;
    		scanf("%d%d", &x, &y);
    		v[x].push_back(y);
    		v[y].push_back(x);
    	}
    }
    
    
    int main(){
    	init();
    	solve();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cxhscst2/p/7160258.html
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