Codeforces 191 C Fools and Roads (tree chain split)

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do the problem mood: do the hdu 5044 after feeling very easy.

Problem Solving Ideas:

First, the tree chain split, the tree is divided into chains, because the final question is all asked, so~ can be linear operation. After the tree chain is split, there will be a lot of chains, but each edge has a number, equivalent to an array of linear operation, so. Add 1 if you go in the U ~ V. Then you can make sum [u] + = 1; sum [v + 1]-= 1; Here if the number of V is large.

At the end of the day, you just have to walk backwards from the back. Get all the results. The idea of a tree-chain split after that is clear.

Code:

#include <iostream> #include <sstream> #include <map> #include <cmath> #include <fstream> #include <queue> #include <vector> #include <sstream> #include <cstring> #include <cstdio > #include <stack> #include <bitset> #include <ctime> #include <string> #include <cctype  > #include <iomanip> #include <algorithm>using namespace std, #define INT __int64#define L (x) (x * 2) #define  R (x) (x * 2 + 1) const int INF = 0X3F3F3F3F; Const double ESP = 0.0000000001; const double PI = ACOs ( -1.0); const INT MOD = 1e9 + 7; const int MY = 1400 + 5; const int MX = 100000 + 5; int n, num, idx, m; int head[mx], sum[mx], ans[mx], p[mx] , Ti[mx], dep[mx], top[mx], siz[mx], son[mx], father[mx]; struct node{int u, v;} E[MX]; struct node{int V, Next;} E[MX*2]; void Addedge (int u, int v) {e[num].v = v; E[num].next = Head[u];    Head[u] = num++; E[NUM].V = u; E[num].next = Head[v]; HEAD[V] = num++;}   void Dfs_find (int u, int fa) { Dep[u] = Dep[fa] + 1;    Siz[u] = 1;    Son[u] = 0;    Father[u] = FA;        for (int i = head[u]; i =-1; i = e[i].next) {int v = E[I].V;        if (v = = FA) continue;        Dfs_find (V, u);        Siz[u] + = Siz[v];    if (Siz[son[u]] < SIZ[V]) son[u] = v;    }}void dfs_time (int u, int fa) {Ti[u] = idx++;    Top[u] = FA;    if (Son[u]) dfs_time (Son[u], top[u]);        for (int i = head[u]; i =-1; i = e[i].next) {int v = E[I].V;        if (v = = Father[u] | | v = = Son[u]) continue;    Dfs_time (V, v); }}void LCA (int u, int v) {while (Top[u]! = Top[v]) {if (Dep[top[u]] < DEP[TOP[V]) swap (U, v)        ;        SUM[TI[U]+1]-= 1;        Sum[ti[top[u]] + = 1;    U = father[top[u]];    } if (Dep[u] > Dep[v]) Swap (U, v);        if (U = v) {Sum[ti[son[u]] + = 1;    SUM[TI[V]+1]-= 1;    }}int Main () {//freopen ("Input.txt", "R", stdin);    int u, v;     while (~SCANF ("%d", &n)) {   num = 0;        memset (head, -1, sizeof (head));        memset (sum, 0, sizeof (sum));            for (int i = 1; i < n; ++i) {scanf ("%d%d", &e[i].u, &AMP;E[I].V);        Addedge (e[i].u, E[I].V);        } dep[1] = siz[0] = 0;        Dfs_find (1, 1);        IDX = 1;        Dfs_time (1, 1);        scanf ("%d", &m);           for (int i = 0; i < m; ++i) {scanf ("%d%d", &u, &v);        LCA (U, v); } for (int i = 1; I <= n; ++i)//First Edge {if (dep[e[i].u] < DEP[E[I].V]) swap (            E[I].U, E[I].V);        P[TI[E[I].U]] = i;            } for (int i = 1; I <= n; ++i)//number of edges in Split {sum[i] + = sum[i-1];        Ans[p[i]] = sum[i];        } printf ("%d", ans[1]);        for (int i = 2; i < n; ++i) printf ("%d", ans[i]);    Puts (""); } return 0;}

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Codeforces 191 C Fools and Roads (tree chain split)