BZOJ 3672[NOI2014]购票（树链剖分+线段树维护凸包+斜率优化） + BZOJ 2402 陶陶的难题II （树链剖分+线段树维护凸包+分数规划+斜率优化）

前言

刚开始看着两道题感觉头皮发麻，后来看看题解，发现挺好理解，只是代码有点长。

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BZOJ 3672[NOI2014]购票

• 中文题面，题意略： BZOJ 3672[NOI2014]购票
• 设f(i)f(i)f(i)表示iii点所花的最小费用，可以写出方程式f(i)=min{ f(j)+pi(disi−disj)+qi }f(i)=min\{\ f(j)+p_i(dis_i-dis_j)+q_i\ \}f(i)=min{ f(j)+pi​(disi​−disj​)+qi​ }其中jjj是iii的祖先且 disi−disj&lt;=lidis_i-dis_j&lt;=l_idisi​−disj​<=li​
• 显然可以斜率优化。那么我们来想想如何在树上做斜率优化。方法就是树链剖分后用dfsdfsdfs序建一颗线段树。线段树的每一个节点上用vectorvectorvector维护这个区间中所有点形成的下凸包。从根往下DPDPDP，在DPDPDP同时维护一个栈来存从根到当前点uuu的链上的点，这些点都有可能转移到当前点。那么只需要在这条链上二分出深度最小且满足 disi−disj&lt;=lidis_i-dis_j&lt;=l_idisi​−disj​<=li​ 的点vvv，在线段树中[dfnv,dfnu][dfn_v,dfn_u][dfnv​,dfnu​]的所有凸包里，二分查询答案就行。树链剖分+线段树+凸包二分，时间复杂度O(nlog3n)O(nlog^3n)O(nlog3n)。

• CODE

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  const int MAXN = 200005;
  const LL inf = 1e18;
  int n, fa[MAXN], S[MAXN], indx;
  int fir[MAXN], to[MAXN], w[MAXN], nxt[MAXN], cnt;
  LL dis[MAXN], f[MAXN], P[MAXN], Q[MAXN], L[MAXN];
  vector<int> t[MAXN<<2];
  int sz[MAXN], hson[MAXN], top[MAXN], dfn[MAXN], seq[MAXN], tmr;
  
  template<typename T>inline void read(T &num) {
      char ch; while((ch=getchar())<'0'||ch>'9');
      for(num=0;ch>='0'&&ch<='9';num=num*10+ch-'0',ch=getchar());
  }
  inline void Addedge(int u, int v, int wt) {
      to[++cnt] = v; nxt[cnt] = fir[u]; fir[u] = cnt; w[cnt] = wt;
  }
  
  inline void dfs(int u) {
      sz[u] = 1;
      for(int i = fir[u]; i; i = nxt[i]) {
          dis[to[i]] = dis[u] + w[i];
          dfs(to[i]); sz[u] += sz[to[i]];
          if(sz[to[i]] > sz[hson[u]]) hson[u] = to[i];
      }
  }
  inline void dfs2(int u, int tp) {
      top[u] = tp; dfn[u] = ++tmr; seq[tmr] = u;
      if(hson[u]) dfs2(hson[u], tp);
      for(int i = fir[u]; i; i = nxt[i])
          if(to[i] != hson[u]) dfs2(to[i], to[i]);
  }
  
  inline double Slope(int i, int j) {
      return (double)(f[i] - f[j]) / (dis[i] - dis[j]);
  }
  
  inline void Modify(int i, int l, int r, int x) {
      int sz = t[i].size();
      while(sz > 1 && Slope(seq[x], t[i][sz-2]) < Slope(t[i][sz-1], t[i][sz-2]))
          t[i].pop_back(), --sz;
      t[i].push_back(seq[x]);
      if(l == r) return;
      int mid = (l + r) >> 1;
      if(x <= mid) Modify(i<<1, l, mid, x);
      else Modify(i<<1|1, mid+1, r, x);
  }
  
  inline LL calc(vector<int>t, int i) {
      int l = 1, r = t.size()-1, mid, pos = 0;
      while(l <= r) {
          mid = (l + r + 1) >> 1;
          if(Slope(t[mid], t[mid-1]) < 1.0*P[i]) pos = mid, l = mid+1;
          else r = mid-1;
      }
      int j = t[pos];
      return f[j] + (dis[i]-dis[j])*P[i] + Q[i];
  }
  
  inline LL Query(int i, int l, int r, int x, int y, int id) {
      if(x <= l && r <= y) return calc(t[i], id);
      int mid = (l + r) >> 1; LL res = inf;
      if(x <= mid) res = min(res, Query(i<<1, l, mid, x, y, id));
      if(y > mid) res = min(res, Query(i<<1|1, mid+1, r, x, y, id));
      return res;
  }
  
  inline void solve(int i, int anc) {
      int u = fa[i]; f[i] = inf;
      while(top[u] != top[anc])
          f[i] = min(f[i], Query(1, 1, n, dfn[top[u]], dfn[u], i)), u = fa[top[u]];
      f[i] = min(f[i], Query(1, 1, n, dfn[anc], dfn[u], i));
  }
  
  inline void dp(int u) {
      S[++indx] = u;
      if(u > 1) {
          int l = 1, r = indx-1, mid;
          while(l < r) {
              mid = (l + r) >> 1;
              if(dis[u] - dis[S[mid]] <= L[u]) r = mid;
              else l = mid+1;
          }
          solve(u, S[l]);
      }
      Modify(1, 1, n, dfn[u]);
      for(int i = fir[u]; i; i = nxt[i]) dp(to[i]);
      --indx;
  }
  
  int main () {
      int type;
      read(n), read(type);
      for(int i = 2, x; i <= n; ++i) {
          read(fa[i]), read(x), read(P[i]), read(Q[i]), read(L[i]);
          Addedge(fa[i], i, x);
      }
      dfs(1); dfs2(1, 1); dp(1);
      for(int i = 2; i <= n; ++i)
          printf("%lld\n", f[i]);
  }
  

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BZOJ 2402 陶陶的难题II

• 题意略： 2402: 陶陶的难题II
• 想想在长度为nnn序列上如何求这个最大值。暴力是O(n2)O(n^2)O(n2)的。
  
  设最终得到最大比值为bestbestbest。那么就有yi+qjxi+pi&lt;=bestyi+qj&lt;=best∗xi+best∗pi(yi−best∗xi)+(qj−best∗pj)&lt;=0\large\begin{aligned}\frac{y_i+q_j}{x_i+p_i}&amp;&lt;=best\\ y_i+q_j&amp;&lt;=best*x_i+best*p_i\\(y_i-best*x_i)+(q_j-best*p_j)&amp;&lt;=0\end{aligned}xi​+pi​yi​+qj​​yi​+qj​(yi​−best∗xi​)+(qj​−best∗pj​)​<=best<=best∗xi​+best∗pi​<=0​对任意i,ji,ji,j都满足，且存在至少一组i,ji,ji,j使等式取等。
• 设等式左边的最大值为fff，那么对于任意取值best′best'best′，有f(best′)&lt;0,    best′&gt;bestf(best′)=0,    best′=bestf(best′)&gt;0,    best′&lt;best\begin{aligned}f(best')&lt;0,&amp;\ \ \ \ best'&gt;best\\f(best')=0,&amp;\ \ \ \ best'=best\\f(best')&gt;0,&amp;\ \ \ \ best'&lt;best\end{aligned}f(best′)<0,f(best′)=0,f(best′)>0,​    best′>best    best′=best    best′<best​
• 显然我们可以二分，每次求最大值fff就行了。那么等式左边的最大值只用分别算(yi−best∗xi)(y_i-best*x_i)(yi​−best∗xi​)和(qj−best∗pj)(q_j-best*p_j)(qj​−best∗pj​)的最大值再加起来。计算最大值时就是用斜率优化，维护一个上凸包。在凸包上二分求最值就行了。
• 而转移到树上就像上一道题一样在dfsdfsdfs序上用线段树维护凸包就行了。
• 这道题buildbuildbuild线段树时需要把子树上传来的两个凸包合并，注意往凸包里加点时要保证xxx坐标递增。
• 时间复杂度O(nlog4n)O(nlog^4n)O(nlog4n)，能过真是奇迹(听说树链跑不满+凸包上的点少?)

• CODE

  #include <bits/stdc++.h>
  using namespace std;
  typedef long long LL;
  const int MAXN = 300005;
  const double inf = 1e16;
  const double eps = 1e-10;
  
  int n; double x[MAXN][2], y[MAXN][2];
  int to[MAXN<<1], nxt[MAXN<<1], fir[MAXN], cnt;
  int sz[MAXN], top[MAXN], fa[MAXN], hson[MAXN], dfn[MAXN], tmr, seq[MAXN], dep[MAXN];
  
  inline void read(int &num) {
  	char ch; while((ch=getchar())<'0'||ch>'9');
  	for(num=0;ch>='0'&&ch<='9';num=num*10+ch-'0',ch=getchar());
  }
  inline void addedge(int u, int v) { to[++cnt] = v; nxt[cnt] = fir[u]; fir[u] = cnt; }
  void dfs(int u, int ff) {
  	fa[u] = ff; sz[u] = 1;
  	dep[u] = dep[fa[u]] + 1;
  	for(int i = fir[u]; i; i = nxt[i])
  		if(to[i] != fa[u]) {
  			dfs(to[i], u), sz[u] += sz[to[i]];
  			if(sz[to[i]] > sz[hson[u]]) hson[u] = to[i];
  		}
  }
  void dfs2(int u, int tp) {
  	top[u] = tp; dfn[u] = ++tmr; seq[tmr] = u;
  	if(hson[u]) dfs2(hson[u], tp);
  	for(int i = fir[u]; i; i = nxt[i])
  		if(to[i] != fa[u] && to[i] != hson[u])
  			dfs2(to[i], to[i]);
  }
  struct SegmentTree {
  	vector<int> vec[MAXN<<2]; bool flg;
  	inline double slope(const int &i, const int &j) {
  		return (y[i][flg]-y[j][flg]) / (x[i][flg]-x[j][flg]);
  	}
  	inline bool Turn_left(const int &i, const int &j, const int &k) {
          register double a = x[j][flg] - x[i][flg], b = y[j][flg] - y[i][flg];
          register double c = x[k][flg] - x[i][flg], d = y[k][flg] - y[i][flg];
          return a*d + eps > b*c;
      }
  	inline void Merge(vector<int> &h, const vector<int> &h1, const vector<int> &h2) {
  		vector<int>::const_iterator i, j;
  		i = h1.begin(), j = h2.begin();
  		int top = 0;
  		while(i != h1.end() || j != h2.end()) { //下面的比较大小就是保证x递增
  			int p = i == h1.end() ? *j++ : j == h2.end() ? *i++ : x[*i][flg] < x[*j][flg] ? *i++ : *j++;
  			while(top >= 2 && Turn_left(h[top-2], h[top-1], p)) h.pop_back(), --top;
  			h.push_back(p), ++top;
  		}
  	}
  	void build(int i, int l, int r) {
  		if(l == r) { vec[i].push_back(seq[l]); return; }
  		register int mid = (l + r) >> 1;
  		build(i<<1, l, mid);
  		build(i<<1|1, mid+1, r);
  		Merge(vec[i], vec[i<<1], vec[i<<1|1]);
  	}
  	inline double calc(const vector<int> &V, const double &now) {
  		register int l = 1, r = V.size()-1, mid, pos = V[0];
  		while(l <= r) {
  			mid = (l + r) >> 1;
  			if(slope(V[mid-1], V[mid]) + eps > now) pos = V[mid], l = mid+1;
  			else r = mid-1;
  		}
  		return y[pos][flg] - now*x[pos][flg];
  	}
  	double Query(const int &i, const int &l, const int &r, const int &L, const int &R, const double &now) {
  		if(L <= l && r <= R) return calc(vec[i], now);
  		int mid = (l + r) >> 1; register double res = -inf;
  		if(L <= mid) res = max(res, Query(i<<1, l, mid, L, R, now));
  		if(R > mid) res = max(res, Query(i<<1|1, mid+1, r, L, R, now));
  		return res;
  	}
  }T[2];
  
  inline double check(int x, int y, const double &now, const bool &flg) {
  	register double res = -inf;
  	register int fx = top[x], fy = top[y];
  	while(fx != fy) {
  		if(dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
  		res = max(res, T[flg].Query(1, 1, n, dfn[top[x]], dfn[x], now));
  		x = fa[fx], fx = top[x];
  	}
  	if(dep[x] < dep[y]) swap(x, y);
  	res = max(res, T[flg].Query(1, 1, n, dfn[y], dfn[x], now));
  	return res;
  }
  inline int lca(int u, int v) {
  	while(top[u] != top[v]) {
  		if(dep[top[u]] > dep[top[v]]) u = fa[top[u]];
  		else v = fa[top[v]];
  	}
  	return dep[u] > dep[v] ? v : u;
  }
  inline bool cmp0(const int &i, const int &j) { return x[i][0] < x[j][0]; }
  inline bool cmp1(const int &i, const int &j) { return x[i][1] < x[j][1]; }
  inline int dcmp(double x) {
  	if(fabs(x) < eps) return 0;
  	if(x > 0) return 1;
  	return -1;
  }
  int main () {
  	read(n); T[1].flg = 1;
  	for(int i = 1; i <= n; ++i) scanf("%lf", &x[i][0]);
  	for(int i = 1; i <= n; ++i) scanf("%lf", &y[i][0]);
  	for(int i = 1; i <= n; ++i) scanf("%lf", &x[i][1]);
  	for(int i = 1; i <= n; ++i) scanf("%lf", &y[i][1]);
  	for(int i = 1, a, b; i < n; ++i)
  		read(a), read(b), addedge(a, b), addedge(b, a);
  	dfs(1, 0); dfs2(1, 1);
  	T[0].build(1, 1, n);
  	T[1].build(1, 1, n);
  	int m, a, b;
  	read(m);
  	while(m--){
  		read(a), read(b);
  		double l = 0, r = 1e8, mid;
  		while(r - l > 1e-5) {
  			mid = (l + r) / 2;
  			if(dcmp(check(a, b, mid, 0) + check(a, b, mid, 1)) >= 0) l = mid;
  			else r = mid;
  		}
  		printf("%.5f\n", l);
  	}
  }
  

• 本人是大常数选手，37192ms37192ms37192ms卡过40000ms40000ms40000ms。