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  • BZOJ 3631 松鼠的新家

    传送门

    分析:

    树链剖分:x->y,将x到y的路径加一,并将x端点的答案-1,最后统计答案。
    树上差分:x->y,x+1,y+1,lca-1,fa[lca]-1,并将x打上标记,最后统计前缀和时将打上标记的点-1.
    两种方法最后都要将终点答案-1.

    code

    差分

    #include<bits/stdc++.h>
    using namespace std;
    namespace IO {
    	template<typename T>
    	inline void read(T &x) {
    		T i = 0, f = 1;
    		char ch = getchar();
    		for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    		if(ch == '-') f = -1, ch = getchar();
    		for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
    		x = i * f;
    	}
    	template<typename T>
    	inline void wr(T x) {
    		if(x < 0) putchar('-'), x = -x;
    		if(x > 9) wr(x / 10);
    		putchar(x % 10 + '0');
    	}
    } using namespace IO;
    
    const int N = 3e5 + 5;
    int n, a[N], ans[N];
    vector<int> G[N];
    int sum[N];
    bool mark[N];
    
    namespace Tree{int idx[N];}
    namespace Tree{
    	int dep[N], son[N], top[N], pos[N], sze[N], tot, fa[N];
    	inline void dfs1(int u, int f){
    		fa[u] = f;
    		dep[u] = dep[f] + 1;
    		for(int e = G[u].size() - 1; e >= 0; e--){
    			int v = G[u][e];
    			if(v == f) continue;
    			dfs1(v, u);
    			sze[u] += sze[v];
    			if(sze[v] > sze[son[u]] || !son[u]) son[u] = v;
    		}
    	}
    	inline void dfs2(int u, int f){
    		if(son[u]){
    			pos[son[u]] = ++tot;
    			idx[tot] = son[u];
    			top[son[u]] = top[u];
    			dfs2(son[u], u);
    		}
    		for(int e = G[u].size() - 1; e >= 0; e--){
    			int v = G[u][e];
    			if(v == f || v == son[u]) continue;
    			pos[v] = ++tot;
    			idx[tot] = v;
    			top[v] = v;
    			dfs2(v, u);
    		}
    	}
    	inline void splitTree(){
    		dfs1(a[1], 0);
    		tot = 1, top[a[1]] = a[1], pos[a[1]] = 1, idx[1] = a[1];
    		dfs2(a[1], 0);
    	}
    	inline int getLca(int u, int v){
    		while(top[u] != top[v]){
    			if(dep[top[u]] < dep[top[v]]) swap(u, v);
    			u = fa[top[u]];
    		}
    		return dep[u] < dep[v] ? u : v;
    	}
    }
    
    inline void getSum(int u, int f){
    	for(int e = G[u].size() - 1; e >= 0; e--){
    		int v = G[u][e];
    		if(v == f) continue;
    		getSum(v, u);
    		sum[u] += sum[v];
    	}
    }
    
    int main() {
    	freopen("h.in", "r" ,stdin);
    	read(n);
    	for(int i = 1; i <= n; i++) read(a[i]);
    	for(int i = 1; i < n; i++){
    		int x, y; read(x), read(y);
    		G[x].push_back(y), G[y].push_back(x);
    	}
    	Tree::splitTree();
    //	dfs(a[1], 0);
    
    	sum[a[1]]++;
    	for(int i = 1; i < n; i++){
    		sum[a[i]]++, sum[a[i + 1]]++;
    		int lca = Tree::getLca(a[i], a[i + 1]);
    		mark[a[i]] = true;
    		sum[lca]--, sum[Tree::fa[lca]]--;		
    	}
    	getSum(a[1], 0);
    	for(int i = 1; i <= n; i++) wr(sum[i] - (i == a[n] ? 1 : 0) - (mark[i] ? 1 : 0)), putchar('
    ');
    	return 0;
    }
    

    树链剖分

    #include<bits/stdc++.h>
    using namespace std;
    namespace IO {
    	template<typename T>
    	inline void read(T &x) {
    		T i = 0, f = 1;
    		char ch = getchar();
    		for(; (ch < '0' || ch > '9') && ch != '-'; ch = getchar());
    		if(ch == '-') f = -1, ch = getchar();
    		for(; ch >= '0' && ch <= '9'; ch = getchar()) i = (i << 3) + (i << 1) + (ch - '0');
    		x = i * f;
    	}
    	template<typename T>
    	inline void wr(T x) {
    		if(x < 0) putchar('-'), x = -x;
    		if(x > 9) wr(x / 10);
    		putchar(x % 10 + '0');
    	}
    } using namespace IO;
    
    const int N = 3e5 + 5;
    int n, a[N], ans[N];
    vector<int> G[N];
    
    namespace Tree{int idx[N];}
    
    namespace Seg{
    	int tree[N << 2], tag[N << 2];
    	inline void upt(int k){
    		tree[k] = tree[k << 1] + tree[k << 1 | 1];
    	}
    	inline void add(int k, int v, int l, int r){
    		tree[k] += (r - l + 1) * v;
    		tag[k] += v;
    	}
    	inline void modify(int k, int l, int r, int x, int y, int v){
    		if(x <= l && r <= y){
    			add(k, v, l, r);
    			return;
    		}
    		int mid = l + r >> 1, lc = k << 1, rc = k << 1 | 1;
    		if(x <= mid) modify(lc, l, mid, x, y, v);
    		if(y > mid) modify(rc, mid + 1, r, x, y, v);
    		upt(k);
    	}
    	inline void pushDown(int k, int l, int r){
    		int mid = l + r >> 1;
    		if(tag[k]){
    			add(k << 1, tag[k], l, mid);
    			add(k << 1 | 1, tag[k], mid + 1, r);
    			tag[k] = 0;
    		}
    	}
    	inline void getAns(int k, int l, int r){
    		if(l == r){
    			ans[Tree::idx[l]] += tree[k];
    			return;
    		}
    		int mid = l + r >> 1;
    		pushDown(k, l, r);
    		getAns(k << 1, l, mid);
    		getAns(k << 1 | 1, mid + 1, r);
    	}
    }
    
    namespace Tree{
    	int dep[N], son[N], top[N], pos[N], sze[N], tot, fa[N];
    	inline void dfs1(int u, int f){
    		fa[u] = f;
    		dep[u] = dep[f] + 1;
    		for(int e = G[u].size() - 1; e >= 0; e--){
    			int v = G[u][e];
    			if(v == f) continue;
    			dfs1(v, u);
    			sze[u] += sze[v];
    			if(sze[v] > sze[son[u]] || !son[u]) son[u] = v;
    		}
    	}
    	inline void dfs2(int u, int f){
    		if(son[u]){
    			pos[son[u]] = ++tot;
    			idx[tot] = son[u];
    			top[son[u]] = top[u];
    			dfs2(son[u], u);
    		}
    		for(int e = G[u].size() - 1; e >= 0; e--){
    			int v = G[u][e];
    			if(v == f || v == son[u]) continue;
    			pos[v] = ++tot;
    			idx[tot] = v;
    			top[v] = v;
    			dfs2(v, u);
    		}
    	}
    	inline void splitTree(){
    		dfs1(a[1], 0);
    		tot = 1, top[a[1]] = a[1], pos[a[1]] = 1, idx[1] = a[1];
    		dfs2(a[1], 0);
    	}
    	inline void pathModify(int u, int v){
    		while(top[u] != top[v]){
    			if(dep[top[u]] < dep[top[v]]) swap(u, v);
    			Seg::modify(1, 1, n, pos[top[u]], pos[u], 1);
    			u = fa[top[u]];
    		}
    		if(dep[u] > dep[v]) swap(u, v);
    		Seg::modify(1, 1, n, pos[u], pos[v], 1);
    	}
    }
    
    inline void dfs(int u, int f){
    	cout<<u<<" ";
    	for(int e = G[u].size() - 1; e >= 0; e--){
    		int v = G[u][e];
    		if(v == f) continue;
    		dfs(v, u);
    	}
    	cout<<endl;
    }
    
    int main() {
    	freopen("h.in", "r" ,stdin);
    	read(n);
    	for(int i = 1; i <= n; i++) read(a[i]);
    	for(int i = 1; i < n; i++){
    		int x, y; read(x), read(y);
    		G[x].push_back(y), G[y].push_back(x);
    	}
    	Tree::splitTree();
    //	dfs(a[1], 0);
    
    	ans[a[1]] = 1;
    	for(int i = 1; i < n; i++) Tree::pathModify(a[i], a[i + 1]), ans[a[i]]--;
    	Seg::getAns(1, 1, n);
    	for(int i = 1; i <= n; i++) wr(ans[i] - (i == a[n] ? 1 : 0)), putchar('
    ');
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/CzYoL/p/7767731.html
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