标签：esc any div algo click span led swap printf
Paths on the tree
Problem Description
bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.**
There are m paths on the tree. bobo would like to pick some paths while any two paths do not share common vertices.**
Find the maximum number of paths bobo can pick.**
Input
The input consists of several tests. For each tests:**
The first line contains n,m (1≤n,m≤105). Each of the following (n - 1) lines contain 2 integers ai,bi denoting an edge between vertices ai and bi (1≤ai,bi≤n). Each of the following m lines contain 2 integers ui,vi denoting a path between vertices ui and vi (1≤ui,vi≤n).**
Output
For each tests:
A single integer, the maximum number of paths.**
Sample Input
3 2
1 2
1 3
1 2
1 3
7 3
1 2
1 3
2 4
2 5
3 6
3 7
2 3
4 5
6 7
Sample Output
1
2
析：n个点的树中有两个点构成的m条路，求每一条路都不包含公共点或边的情况下最多的路的条数，用LCA倍增算法，求出每条路的两个点的LCA，并对其LCA按深度从大到小排序，先取深度较大的边，这样并不会对浅边产生影响，但要维护其LCA下的所有子节点，对其进行标记，避免重复
#include <vector> #include <queue> #include <math.h> #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #define ll long long #define ull unsigned ll #define Inf 0x7fffffff #define maxn 100005 #define maxm 100005 #define P pair<int, int> using namespace std; const int N = 300005; int t, n, m, len, ans; int vis[maxn], deep[maxn]; int x, y, mark[maxn], f[maxn][30]; vector<int>V[maxn]; struct Node{ int x, y, p; bool operator<(const Node &a)const{ return deep[p] > deep[a.p]; } }g[maxn]; void init(){ for(int i = 1; i <= n; i ++){ vis[i] = deep[i] = mark[i] = 0; fill(f[i], f[i]+30, 0); V[i].clear(); } } void dfs(int u){ vis[u] = 0; for(int i = 0; i < V[u].size(); i ++){ int v = V[u][i]; if(f[u][0] == v) continue; if(!vis[v]){ deep[v] = deep[u]+1; f[v][0] = u; dfs(v); } } } void cal(){ for(int i = 1; (1<<i) <= n; i ++){ for(int j = 1; j <= n; j ++){ f[j][i] = f[f[j][i-1]][i-1]; } } } int LCA(int x, int y){ if(deep[x] < deep[y]) swap(x, y); int j, i = 0; while((1<<(i+1)) <= deep[x]) i ++; for(j = i; j+1; j --){ if(deep[x]-(1<<j) >= deep[y]){ x = f[x][j]; } } if(x == y) return x; for(j = i; j+1; j --){ if(f[x][j] != f[y][j]){ x = f[x][j]; y = f[y][j]; } } return f[x][0]; } void solve(int u){ mark[u] = 1; for(int i = 0; i < V[u].size(); i ++){ int v = V[u][i]; if(deep[u]+1 == deep[v] && !mark[v]){ solve(v); } } } int main(){ while(~scanf("%d%d", &n, &m)){ init(); for(int i = 1; i < n; i ++){ scanf("%d%d", &x, &y); V[x].push_back(y); V[y].push_back(x); } deep[1] = f[1][0] = 1; dfs(1), cal(); for(int i = 0; i < m; i ++){ scanf("%d%d", &g[i].x, &g[i].y); g[i].p = LCA(g[i].x, g[i].y); } int res = 0; sort(g, g+m); for(int i = 0; i < m; i ++){ if(!mark[g[i].x] && !mark[g[i].y]){ res ++; solve(g[i].p); } } printf("%d\n", res); } return 0; }
标签：esc any div algo click span led swap printf
原文地址：https://www.cnblogs.com/microcodes/p/12732276.html
