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  • P5838 [USACO19DEC]Milk Visits G

    题解

    并不需要什么高级数据结构
    用树链剖分维护
    对于每种颜色开个 ( ext{vector}),然后把是这种颜色的点的 ( ext{dfs}) 序加进来排序
    对于 ([dfn[top[x]],dfn[x]]) 这一区间问有没有某种颜色
    相当于问某种颜色有没有至少一个在这个区间内
    直接二分查 ( ext{vector}) 即可

    (Code)

    #include<cstdio>
    #include<vector>
    #include<algorithm>
    using namespace std;
    
    const int N = 1e5 + 5;
    int n, m, a[N], h[N], ans[N];
    vector<int> col[N];
    
    struct edge{int to, nxt;}e[N << 1];
    inline void add(int x, int y)
    {
    	static int tot = 0;
    	e[++tot] = edge{y, h[x]}, h[x] = tot;
    }
    
    int siz[N], son[N], dfn[N], fa[N], top[N], dep[N];
    void dfs1(int x)
    {
    	siz[x] = 1, dep[x] = dep[fa[x]] + 1;
    	for(register int i = h[x]; i; i = e[i].nxt)
    	{
    		int v = e[i].to;
    		if (v == fa[x]) continue;
    		fa[v] = x, dfs1(v), siz[x] += siz[v];
    		if (siz[v] > siz[son[x]]) son[x] = v;
    	}
    }
    void dfs2(int x)
    {
    	static int dfc = 0;
    	dfn[x] = ++dfc;
    	if (son[x]) top[son[x]] = top[x], dfs2(son[x]);
    	for(register int i = h[x]; i; i = e[i].nxt)
    	{
    		int v = e[i].to;
    		if (v == fa[x] || v == son[x]) continue;
    		top[v] = v, dfs2(v);
    	}
    }
    
    inline int search(int x, int y, int c)
    {
    	int l = 0, r = col[c].size() - 1, mid, ll = -1, rr = -1;
    	while (l <= r)
    	{
    		mid = (l + r) >> 1;
    		if (col[c][mid] >= x) ll = mid, r = mid - 1;
    		else l = mid + 1;
    	}
    	if (ll == -1) return 0;
    	l = ll, r = col[c].size() - 1;
    	while (l <= r)
    	{
    		mid = (l + r) >> 1;
    		if (col[c][mid] <= y) rr = mid, l = mid + 1;
    		else r = mid - 1;
    	}
    	if (rr == -1) return 0;
    	return 1;
    }
    
    inline int solve(int x, int y, int c)
    {
    	int fx = top[x], fy = top[y];
    	while (fx ^ fy)
    	{
    		if (dep[fx] > dep[fy])	
    		{
    			if (search(dfn[fx], dfn[x], c)) return 1;
    			x = fa[fx], fx = top[x];
    		}
    		else{
    			if (search(dfn[fy], dfn[y], c)) return 1;
    			y = fa[fy], fy = top[y];
    		}
    	}
    	if (dep[x] < dep[y]) return search(dfn[x], dfn[y], c);
    	return search(dfn[y], dfn[x], c);
    }
    
    int main()
    {
    	scanf("%d%d", &n, &m);
    	for(register int i = 1; i <= n; i++) scanf("%d", a + i);
    	for(register int i = 1, u, v; i < n; i++) scanf("%d%d", &u, &v), add(u, v), add(v, u);
    	dfs1(1), top[1] = 1, dfs2(1);
    	for(register int i = 1; i <= n; i++) col[a[i]].push_back(dfn[i]);
    	for(register int i = 1; i <= n; i++) sort(col[i].begin(), col[i].end());
    	int cnt = 0;
    	for(int u, v, w; m; --m)
    	{
    		scanf("%d%d%d", &u, &v, &w);
    		if (solve(u, v, w)) ans[++cnt] = 1;
    		else ans[++cnt] = 0;
    	}
    	for(register int i = 1; i <= cnt; i++) printf("%d", ans[i]);
    }
    
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  • 原文地址:https://www.cnblogs.com/leiyuanze/p/14316972.html
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