题意是将一颗树cut一条边成两棵树,求生成的两棵树之间的连接边的数目。
对于每条不是树上的边<a, b>, a节点加1, b节点加1,LAC(a, b)减 2,对每颗子树求和
#include <iostream> #include <algorithm> #include <vector> #include <cstring> #include <cstdio> #define N 20010 using namespace std; struct node{ vector<int> adj; vector<int> quary; int add; node(){ clear(); } void clear(){ adj.clear(); quary.clear(); add = 0; } } p[N]; int fa[N], visit[N]; int find(int x) { return x==fa[x] ? x : (fa[x] = find(fa[x])); } void LCA(int x, int f){ for(int i=0; i<p[x].adj.size(); ++i) { if(p[x].adj[i] == f) continue; LCA(p[x].adj[i], x); fa[ p[x].adj[i] ] = x; } visit[x] = 1; for(int i=0; i<p[x].quary.size(); ++i) { if(visit[ p[x].quary[i] ]) { ++ p[x].add; ++ p[ p[x].quary[i] ].add; p[ find( p[x].quary[i]) ].add -= 2; } } } int re; void dfs(int x, int f){ for(int i=0; i<p[x].adj.size(); ++i) { if(p[x].adj[i] == f) continue; dfs(p[x].adj[i], x); p[x].add += p[ p[x].adj[i] ].add; } if(x != 0) { re = min(re, p[x].add); } } int main(int argc, char* argv[]){ int t, ca = 1; scanf("%d", &t); while(t--){ int n, m; scanf("%d%d", &n, &m); for(int i=0; i<n; ++i){ p[i].clear(); fa[i] = i; } memset(visit, 0, sizeof(visit)); for(int i=0; i<n-1; ++i) { int a, b; scanf("%d%d", &a, &b); --a, --b; p[a].adj.push_back(b); p[b].adj.push_back(a); } for(int i=n-1; i<m; ++i) { int a, b; scanf("%d%d", &a, &b); --a, --b; p[a].quary.push_back(b); p[b].quary.push_back(a); } LCA(0, -1); re = INT_MAX; dfs(0, -1); printf("Case #%d: %d\n", ca++, re + 1); } return 0; }