「CJOJ2722」Ping

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Sample Input

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5 4
2 1
5 3
3 1
4 3
2
2 4
3 2

Sample Output

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1

Hint

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题解

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这道题目显然把每一段查询的路径求出来（即求LCA），然后将LCA排序，最后算一遍就行了

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
#define re register
using namespace std;
inline int gi(){
	int sum=0,f=1;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return sum*f;
}
const int maxn=100010;
struct gjh{
	int to,nxt;
}e[maxn<<1],e1[maxn<<3];
int dfs[maxn];
struct node{
	int x,y,z;
	bool operator <(node b)const{
		return dfs[z]>dfs[b.z];
	}
}lca[maxn<<2];
int front[maxn],cnt,p[maxn],k,bj[maxn];
int fa[maxn],lcah[maxn],cou;
void Add(int u,int v){
	e[++cnt]=(gjh){v,front[u]};front[u]=cnt;
	e[++cnt]=(gjh){u,front[v]};front[v]=cnt;
}
void Ad(int x,int y){
	e1[++cnt]=(gjh){y,p[x]};p[x]=cnt;
	e1[++cnt]=(gjh){x,p[y]};p[y]=cnt;
}
int find(int x){
	if(lcah[x]!=x)lcah[x]=find(lcah[x]);
	return lcah[x];
}
void tarjan(int u){
	bj[u]=1;lcah[u]=u;dfs[u]=++cnt;
	for(int i=front[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v!=fa[u]){
			fa[v]=u;
			tarjan(v);
			lcah[v]=u;
		}
	}
	for(int i=p[u];i;i=e1[i].nxt)
		if(bj[e1[i].to])
			lca[++cou]=(node){u,e1[i].to,find(e1[i].to)};
}
int cmp(node a,node b){
	return a.z>b.z;
}
int a[maxn<<2],ans,flag[maxn];
void dfs1(int u){
	flag[u]=1;
	for(int i=front[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v!=fa[u])dfs1(v);
	}
}
int main(){
	int i,j,n,m;
	n=gi();m=gi();
	for(i=1;i<=m;i++){
		int u=gi(),v=gi();Add(u,v);
	}
	k=gi();cnt=0;
	for(i=1;i<=k;i++){
		int x=gi(),y=gi();
		Ad(x,y);
	}
	fa[1]=1;cnt=0;
	tarjan(1);
	sort(lca+1,lca+cou+1);
	for(i=1;i<=cou;i++)
		if(!flag[lca[i].x] && !flag[lca[i].y]){
			a[++ans]=lca[i].z;
			dfs1(lca[i].z);
		}
	printf("%d\n",ans);
	return 0;
}