Description
Yixght is a manager of the company called SzqNetwork(SN). Now she's very worried because she has just received a bad news which denotes that DxtNetwork(DN), the SN's business rival, intents to attack the network of SN. More unfortunately, the original network of SN is so weak that we can just treat it as a tree. Formally, there are N nodes in SN's network, N-1 bidirectional channels to connect the nodes, and there always exists a route from any node to another. In order to protect the network from the attack, Yixght builds M new bidirectional channels between some of the nodes.
As the DN's best hacker, you can exactly destory two channels, one in the original network and the other among the M new channels. Now your higher-up wants to know how many ways you can divide the network of SN into at least two parts.
Input
The first line of the input file contains two integers: N (1 ≤ N ≤ 100 000), M (1 ≤ M ≤ 100 000) — the number of the nodes and the number of the new channels.
Following N-1 lines represent the channels in the original network of SN, each pair (a,b) denote that there is a channel between node a and node b.
Following M lines represent the new channels in the network, each pair (a,b) denote that a new channel between node a and node b is added to the network of SN.
Output
Output a single integer — the number of ways to divide the network into at least two parts.
Sample Input
4 1
1 2
2 3
1 4
3 4
Sample Output
3
题目链接:http://poj.org/problem?id=3417
解题报告
树上差分，可以差出一个结点(根节点除外)上方的边被多少条新边取代.
那么没有被代取的做出贡献为m，能被一条边取代的贡献为1，能被多条边取代的贡献为0.
dfs统计答案.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#define ll long long
#define BIG 200011
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
ll ans;
int n,m,tot;
int x,y,z,dfn_num;
int nxt[BIG],las[BIG],to[BIG],sign[BIG],f[BIG],sz[BIG],dep[BIG],xu[BIG],top[BIG];
inline int read(){
char c=getchar();
while(c>'9'||c<'0')c=getchar();
int data=0;
while(c<='9'&&c>='0')data=(data<<1)+(data<<3)+c-48,c=getchar();
return data;
}
inline void add(int x,int y){
nxt[++tot]=las[x];
las[x]=tot;
to[tot]=y;
return;
}
inline void dfs1(int now){
++sz[now];
for(register int e=las[now];e;e=nxt[e])
if(to[e]!=f[now]){
dep[to[e]]=dep[now]+1;
f[to[e]]=now;
dfs1(to[e]);
sz[now]+=sz[to[e]];
}
return;
}
inline void dfs2(int now,int chain){
xu[now]=++dfn_num,top[now]=chain;
register int i=0;
for(register int e=las[now];e;e=nxt[e])
if(to[e]!=f[now]&&sz[to[e]]>sz[i])i=to[e];
if(!i)return;
dfs2(i,chain);
for(register int e=las[now];e;e=nxt[e])
if(to[e]!=f[now]&&to[e]!=i)dfs2(to[e],to[e]);
return;
}
inline int lca(int x,int y){
for(;top[x]!=top[y];dep[top[x]]>dep[top[y]]?x=f[top[x]]:y=f[top[y]]);
return dep[x]<dep[y]?x:y;
}
inline void dfs3(int now){
for(register int e=las[now];e;e=nxt[e])
if(to[e]!=f[now]){
dfs3(to[e]);
sign[now]+=sign[to[e]];
}
if(now==1)return;
if(sign[now]==1)++ans;
if(sign[now]==0)ans+=m;
return;
}
int main(){
n=read(),m=read();
FOR(i,1,n-1){
x=read(),y=read();
add(x,y);add(y,x);
}
dep[1]=1;dfs1(1);dfs2(1,1);
for(register int i=1;i<=m;++i){
x=read(),y=read();
++sign[x];++sign[y];sign[lca(x,y)]-=2;
}
dfs3(1);
cout<<ans<<endl;
return 0;
}