[HDU5452]Minimum Cut

Problem Description
Given a simple unweighted graph G(an undirected graph containing no loops nor multiple edges) with n nodes and m edges. Let Tbe a spanning tree of G .
We say that a cut in G respects T  if it cuts just one edges of T. Since love needs good faith and hypocrisy return for only grief, you should find the minimum cut of graph G

respecting the given spanning tree T.

Input
The input contains several test cases.
The first line of the input is a single integer t (1≤t≤5) which is the number of test cases.
Then t test cases follow.
Each test case contains several lines.
The first line contains two integers n (2≤n≤20000) and m (n−1≤m≤200000).
The following n−1 lines describe the spanning tree T and each of them contains two integers u and v corresponding to an edge.
Next m−n+1 lines describe the undirected graph G and each of them contains two integers u and v
corresponding to an edge which is not in the spanning tree T.
 
Output
For each test case, you should output the minimum cut of graph G respecting the given spanning tree T.
 
Sample Input
1
4 5
1 2
2 3
3 4
1 3
1 4
 
Sample Output
Case #1: 2
 
Source
2015 ACM/ICPC Asia Regional Shenyang Online
 
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题意：

给定一张n个点，m条边的无向图，并给出它的一棵生成树 。你需要删去一些边，使得剩下的图不连通，且删去的边中必须恰好包含一条树边。 求最少删去的边数。

题解：

枚举树上的每一条边，将其删掉后，树分裂成了两部分，必须删掉所有横跨两部分的非树边。
对于每条非树边，给树上对应路径上的边的答案都加 1。

实现路径覆盖次数加1：在树上做差分。对于路径u-v，再点u和点v上+1，在其LCA上-2。

#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int INF=0x7f7f7f7f;const int N=20010; const int M=400010; int T, n, m; void Getin( int &shu ) {     char c; int f=1; shu=0;     for( c=getchar(); c<'0' || c>'9'; c=getchar() ) if( c=='-' ) f=-1;     for( ; c>='0' && c<='9'; c=getchar() ) shu=shu*10+c-'0';     shu*=f; } int fir[N], ecnt; struct nodes{ int e, next; }edge[M]; void Link( int s, int e ) {     edge[++ecnt].e=e; edge[ecnt].next=fir[s]; fir[s]=ecnt;     edge[++ecnt].e=s; edge[ecnt].next=fir[e]; fir[e]=ecnt; }int fa[N], son[N], dep[N], siz[N];void DFS1( int r, int f, int d ) {dep[r]=d; siz[r]=1;for( int i=fir[r]; i; i=edge[i].next )if( edge[i].e!=f ) {DFS1( edge[i].e, r, d+1 );siz[r]+=siz[ edge[i].e ];if( siz[ edge[i].e ]>siz[ son[r] ] ) son[r]=edge[i].e;fa[ edge[i].e ]=r;}}int top[N];void DFS2( int r, int f ) {if( son[r] && son[r]!=r ) top[ son[r] ]=top[r], DFS2( son[r], r );for( int i=fir[r]; i; i=edge[i].next )if( edge[i].e!=f && edge[i].e!=son[r] ) {top[ edge[i].e ]=edge[i].e;DFS2( edge[i].e, r );}}int LCA( int p1, int p2 ) {while( top[p1]!=top[p2] ) {if( dep[ top[p1] ]>dep[ top[p2] ] ) p1=fa[ top[p1] ];else p2=fa[ top[p2] ];}return dep[p1]<dep[p2] ? p1 : p2;}int cnt[N];void DFS( int r, int f ) {for( int i=fir[r]; i; i=edge[i].next )if( edge[i].e!=f ) {DFS( edge[i].e, r );cnt[r]+=cnt[ edge[i].e ];}}void Reset() {ecnt=0;memset( fir, 0, sizeof fir );memset( cnt, 0, sizeof cnt );memset( son, 0, sizeof son );}int s, e, cas;int main() {for( Getin(T); T; T-- ) {Reset();Getin(n); Getin(m);for( int i=1; i<n; i++ ) {Getin(s); Getin(e);Link( s, e );}DFS1( 1, -1, 1 ); top[1]=1; DFS2( 1, -1 ); for( int i=n; i<=m; i++ ) {Getin(s); Getin(e);int f=LCA( s, e );cnt[s]++; cnt[e]++;cnt[f]-=2;}DFS( 1, -1 );int ans=INF;for( int i=2; i<=n; i++ )ans=min( ans, cnt[i]+1 );printf( "Case #%d: %d\n", ++cas, ans );}return 0;}