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  • NOIP 2016 天天爱跑步 (luogu 1600 & uoj 261)

    题目传送门

    　　传送点I

    　　传送点II

    题目大意

    　　(此题目不需要大意，我认为它已经很简洁了)

    　　显然线段树合并(我也不知道哪来这么多显然)

    　　考虑将每条路径拆成两条路径 s -> lca 和 t -> lca 。

    　　对于前一种路径上的某一点i，希望在时刻 w[i] 经过它，那么就有

     dep[s] - dep[i] = w[i] 

    　　移项可得:

     dep[s] = w[i] + dep[i] 

    　　然后发现dep[s]可以被看做已知条件，那么根据常用套路，在点s将线段树dep[s]处的值 + 1，在lca处还原。

    　　回溯的过程中通过 w[i] + dep[i] 查答案就好了。

    　　对于后一种路径，考虑在点t的时候将线段树的某个位置 + 1，在lca计算答案之前还原。那么就有:

     dep[s] + dep[i] - 2 * dep[lca] = w[i] 

    　　然后移项得到:

     dep[s] - 2 * dep[lca] = w[i] - dep[i] 

    　　继续扔进线段树里维护和查询。

    　　(前年做这道题的时候完全没有思路，去年发现原来这么水。。)

    Code

      1 /**
  2  * luogu
  3  * Problem#1600
  4  * Accepted
  5  * Time: 6321ms
  6  * Memory: 172394k
  7  */
  8 #include <bits/stdc++.h>
  9 using namespace std;
 10 #define smin(_a, _b) _a = min(_a, _b)
 11 #define smax(_a, _b) _a = max(_a, _b)
 12 #define fi first
 13 #define sc second
 14 typedef pair<int, int> pii;
 15 typedef bool boolean;
 16 template<typename T>
 17 inline void readInteger(T& u) {
 18     static char x;
 19     while(!isdigit(x = getchar()));
 20     for(u = x - '0'; isdigit(x = getchar()); u = u * 10 + x - '0');
 21 }
 22 
 23 typedef class SegTreeNode {
 24     public:
 25         int val;
 26         SegTreeNode *l, *r;
 27         
 28         SegTreeNode(int val = 0, SegTreeNode* l = NULL, SegTreeNode* r = NULL):val(val), l(l), r(r) {        }
 29         
 30         inline void pushUp() {
 31             val = l->val + r->val;
 32         }
 33 }SegTreeNode;
 34 
 35 #define LIMIT 2000000
 36 SegTreeNode pool[LIMIT + 1];
 37 SegTreeNode *top = pool;
 38 SegTreeNode null(0, &null, &null);
 39 
 40 SegTreeNode* newnode() {
 41     if(top >= pool + LIMIT)
 42         return new SegTreeNode(0, &null, &null);
 43     *top = SegTreeNode(0, &null, &null);
 44     return top++;
 45 }
 46 
 47 #define null &null
 48 
 49 void merge(SegTreeNode*& a, SegTreeNode* b) {
 50     if(a == null) {
 51         a = b;
 52         return;
 53     }
 54     if(b == null)    return;
 55     a->val += b->val;
 56     merge(a->l, b->l);
 57     merge(a->r, b->r);
 58 }
 59 
 60 typedef class SegTree {
 61     public:
 62         SegTreeNode* root;
 63         
 64         SegTree():root(null) {        }
 65         
 66         void update(SegTreeNode*& node, int l, int r, int idx, int val) {
 67             if(node == null)
 68                 node = newnode();
 69             if(l == idx && r == idx) {
 70                 node->val += val;
 71                 return;
 72             }
 73             int mid = (l + r) >> 1;
 74             if(idx <= mid)
 75                 update(node->l, l, mid, idx, val);
 76             else
 77                 update(node->r, mid + 1, r, idx, val);
 78             node->pushUp();
 79         }
 80         
 81         int query(SegTreeNode*& node, int l, int r, int idx) {
 82             if(node == null)    return 0;
 83             if(l == idx && r == idx)
 84                 return node->val;
 85             int mid = (l + r) >> 1;
 86             if(idx <= mid)
 87                 return query(node->l, l, mid, idx);
 88             return query(node->r, mid + 1, r, idx);
 89         }
 90 }SegTree;
 91 
 92 typedef class Query {
 93     public:
 94         int s;
 95         int t;
 96         int lca;
 97         
 98         Query(int s = 0, int t = 0, int lca = 0):s(s), t(t), lca(lca) {        } 
 99 }Query;
100 
101 int n, m;
102 int* wss;
103 Query* qs;
104 vector<int> *q;
105 vector<int> *g;
106 
107 inline void init() {
108     readInteger(n);
109     readInteger(m);
110     wss = new int[(n + 1)];
111     qs = new Query[(m + 1)];
112     q = new vector<int>[(n + 1)];
113     g = new vector<int>[(n + 1)];
114     for(int i = 1, u, v; i < n; i++) {
115         readInteger(u);
116         readInteger(v);
117         g[u].push_back(v);
118         g[v].push_back(u);
119     }
120     for(int i = 1; i <= n; i++)
121         readInteger(wss[i]);
122     for(int i = 1, s, t; i <= m; i++) {
123         readInteger(s);
124         readInteger(t);
125         qs[i] = Query(s, t, 0);
126         q[s].push_back(i);
127         if(s != t)
128             q[t].push_back(i);
129     }
130 }
131 
132 int* f;
133 int* dep;
134 int find(int x) {
135     return (f[x] == x) ? (x) : (f[x] = find(f[x])); 
136 }
137 
138 void tarjan(int node, int fa) {
139     f[node] = node;
140     dep[node] = dep[fa] + 1;
141     for(int i = 0; i < (signed)g[node].size(); i++) {
142         int& e = g[node][i];
143         if(e == fa) continue;
144         tarjan(e, node);
145         f[e] = node;
146     }
147     for(int i = 0; i < (signed)q[node].size(); i++) {
148         int id = q[node][i];
149         if(f[qs[id].s] && f[qs[id].t] && !qs[id].lca)
150             qs[id].lca = find(f[(qs[id].s == node) ? (qs[id].t) : (qs[id].s)]);
151     }
152 }
153 
154 int *ans;
155 vector<int>* ls;
156 SegTreeNode* dfs1(int node, int fa) {
157     SegTree st;
158     for(int i = 0; i < (signed)q[node].size(); i++) {
159         Query &aq = qs[q[node][i]];
160         if(aq.s == node) {
161             st.update(st.root, 1, n, dep[aq.s], 1);
162             ls[aq.lca].push_back(q[node][i]);
163         }
164     }
165     for(int i = 0; i < (signed)g[node].size(); i++) {
166         int& e = g[node][i];
167         if(e == fa)    continue;
168         merge(st.root, dfs1(e, node));
169     }
170     if(dep[node] + wss[node] <= n)
171         ans[node] = st.query(st.root, 1, n, dep[node] + wss[node]);
172     else
173         ans[node] = 0;
174     for(int i = 0; i < (signed)ls[node].size(); i++)
175         st.update(st.root, 1, n, dep[qs[ls[node][i]].s], -1);
176     ls[node].clear();
177     return st.root;
178 }
179 
180 SegTreeNode* dfs2(int node, int fa) {
181     SegTree st;
182     for(int i = 0; i < (signed)q[node].size(); i++) {
183         Query &aq = qs[q[node][i]];
184         if(aq.t == node) {
185             st.update(st.root, -n, n, dep[aq.s] - 2 * dep[aq.lca], 1);
186             ls[aq.lca].push_back(q[node][i]);
187         }
188     }
189     for(int i = 0; i < (signed)g[node].size(); i++) {
190         int& e = g[node][i];
191         if(e == fa)    continue;
192         merge(st.root, dfs2(e, node));
193     }
194     for(int i = 0; i < (signed)ls[node].size(); i++)
195         st.update(st.root, -n, n, dep[qs[ls[node][i]].s] - 2 * dep[qs[ls[node][i]].lca], -1);
196     ans[node] += st.query(st.root, -n, n, wss[node] - dep[node]);
197     return st.root;
198 }
199 
200 inline void solve() {
201     f = new int[(n + 1)];
202     dep = new int[(n + 1)];
203     dep[0] = 0;
204     memset(f, 0, sizeof(int) * (n + 1));
205     tarjan(1, 0);
206     ans = new int[(n + 1)];
207     ls = new vector<int>[(n + 1)];
208     dfs1(1, 0);
209     top = pool;
210     for(int i = 1; i <= n; i++)
211         assert(ls[i].empty());
212     dfs2(1, 0);
213     for(int i = 1; i <= n; i++)
214         printf("%d ", ans[i]);
215 }
216 
217 int main() {
218     init();
219     solve();
220     return 0;
221 }
    Segment Tree

    　　线段树不优秀，跑得太慢了。ccf老年机应该会跑T。

    　　我们发现，这个本质上是在dfs序某些位置某一下标修改一个值，然后询问区间某一下标的和。

    　　这个完全可以用前缀和相减，而没必要出动线段树。

    　　比较简单的写法就是：访问到一个点，记录一下它询问的下标的值，然后再递归它的子树，最后加上差值。

    Code

      1 /**
  2  * uoj
  3  * Problem#261
  4  * Accepted
  5  * Time: 1236ms
  6  * Memory: 59108k
  7  */
  8 #include <iostream>
  9 #include <cstdlib>
 10 #include <cstdio>
 11 #include <vector>
 12 using namespace std;
 13 typedef bool boolean;
 14 
 15 const int N = 3e5 + 3, N2 = N << 1;
 16 
 17 #define pii pair<int, int>
 18 #define fi first
 19 #define sc second
 20 
 21 template <typename T>
 22 void pfill(T* ps, const T* ped, T val) {
 23     for ( ; ps != ped; *ps = val, ps++);
 24 }
 25 
 26 template <typename T>
 27 class MapManager {
 28     public:
 29         int* h;
 30         vector< pair<T, int> > vs;
 31 
 32         MapManager() {    }
 33         MapManager(int n) {
 34             h = new int[(n + 1)];
 35             pfill(h, h + n + 1, -1);
 36         }
 37 
 38         void insert(int p, T dt) {
 39             vs.push_back(make_pair(dt, h[p]));
 40             h[p] = (signed) vs.size() - 1;
 41         }
 42 
 43         pair<T, int>& operator [] (int p) {
 44             return vs[p];
 45         }
 46 
 47 };
 48 
 49 int n, m;
 50 int *dep;
 51 int *wts, *res;
 52 int *lcas, *us, *vs;
 53 boolean *exi;
 54 MapManager<int> g;
 55 MapManager<pii> qlca;    // fi: another node. sc: id
 56 MapManager<int> as, rs;
 57 MapManager<pii> ms;    // sc: val
 58 
 59 inline void init() {
 60     scanf("%d%d", &n, &m);
 61     wts = new int[(n + 1)];
 62     res = new int[(n + 1)];
 63     exi = new boolean[(m + 1)];
 64     pfill(res + 1, res + n + 1, 0);
 65     pfill(exi, exi + m + 1, true);
 66     g = MapManager<int>(n);
 67     for (int i = 1, u, v; i < n; i++) {
 68         scanf("%d%d", &u, &v);
 69         g.insert(u, v), g.insert(v, u);
 70     }
 71     for (int i = 1; i <= n; i++)
 72         scanf("%d", wts + i);
 73     us = new int[(n + 1)];
 74     vs = new int[(n + 1)];
 75     lcas = new int[(n + 1)];
 76     qlca = MapManager<pii>(n);
 77     for (int i = 1, u, v; i <= m; i++) {
 78         scanf("%d%d", &u, &v);
 79         qlca.insert(u, pii(v, i));
 80         qlca.insert(v, pii(u, i));
 81         us[i] = u, vs[i] = v;
 82     }
 83 }
 84 
 85 int *uf;
 86 boolean *vis;
 87 int dfs_clock;
 88 
 89 int find(int x) {
 90     return (uf[x] == x) ? (x) : (uf[x] = find(uf[x]));
 91 }
 92 
 93 void tarjan(int p, int fa, int dp) {
 94     vis[p] = true, dep[p] = dp;
 95     for (int i = g.h[p], e; ~i; i = g[i].sc) {
 96         if ((e = g[i].fi) == fa)
 97             continue;
 98         tarjan(e, p, dp + 1);
 99         uf[find(e)] = p;
100     }
101 
102     for (int i = qlca.h[p]; ~i; i = qlca[i].sc) {
103         pii d = qlca[i].fi;
104         if (vis[d.fi] && exi[d.sc]) {
105             exi[d.sc] = false;
106             lcas[d.sc] = find(d.fi);
107         }
108     }
109 }
110 
111 int bucket[N2];
112 
113 void put(int p, int val) {
114     (p < 0) ? (p += N) : (0);
115     bucket[p] += val;
116 }
117 
118 int get(int p) {
119     (p < 0) ? (p += N) : (0);
120     return bucket[p];
121 }
122 
123 void dfs1(int p, int fa) {
124     int tmp = get(wts[p] + dep[p]);
125     for (int i = as.h[p]; ~i; i = as[i].sc)
126         put(as[i].fi, 1);
127     for (int i = g.h[p], e; ~i; i = g[i].sc) {
128         if ((e = g[i].fi) == fa)
129             continue;
130         dfs1(e, p);
131     }
132     res[p] += get(wts[p] + dep[p]) - tmp;
133     for (int i = rs.h[p]; ~i; i = rs[i].sc)
134         put(rs[i].fi, -1);
135 }
136 
137 void dfs2(int p, int fa) {
138     int tmp = get(wts[p] - dep[p]);
139     for (int i = ms.h[p]; ~i; i = ms[i].sc)
140         put(ms[i].fi.fi, ms[i].fi.sc);
141     for (int i = g.h[p], e; ~i; i = g[i].sc) {
142         if ((e = g[i].fi) == fa)
143             continue;
144         dfs2(e, p);
145     }
146     res[p] += get(wts[p] - dep[p]) - tmp;
147 }
148 
149 inline void solve() {
150     uf = new int[(n + 1)];
151     dep = new int[(n + 1)];
152     vis = new boolean[(n + 1)];
153     for (int i = 1; i <= n; i++)
154         uf[i] = i;
155     pfill(vis + 1, vis + n + 1, false);
156     tarjan(1, 0, 1);
157     
158     delete[] vis;
159     delete[] exi;
160 
161     as = MapManager<int>(n);
162     rs = MapManager<int>(n);
163     ms = MapManager<pii>(n);
164 
165     for (int i = 1; i <= m; i++) {
166         int u = us[i], v = vs[i], g = lcas[i];
167         as.insert(u, dep[u]);
168         rs.insert(g, dep[u]);
169         ms.insert(v, pii(dep[u] - 2 * dep[g], 1));
170         ms.insert(g, pii(dep[u] - 2 * dep[g], -1));
171     }
172 
173     dfs1(1, 0);
174     dfs2(1, 0);
175     for (int i = 1; i <= n; i++)
176         printf("%d ", res[i]);
177 }
178 
179 int main() {
180     init();
181     solve();
182     return 0;
183 }
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  • 原文地址：https://www.cnblogs.com/yyf0309/p/7749541.html
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