題意：有N台電腦，他們之間有M條無向邊。
然後詢問，每次在他們之間加一條邊，剩余的橋有多少。
思路：其實這題都不需要縮點了。。
直接記錄每條橋的位置，然後每次詢問進行一次LCA，當詢問到橋時，橋數減1，下次詢問就不會再計數了。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 2505
#define inf 1<<28
#define LL(x) ( x << 1 )
#define RR(x) ( x << 1 | 1 )
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define PII pair<int,int>
using namespace std;
#define M 500005
#define N 200005
#define stop system("pause")
inline void RD(int &ret) {
char c;
do {
c = getchar();
} while(c < '0' || c > '9') ;
ret = c - '0';
while((c=getchar()) >= '0' && c <= '9')
ret = ret * 10 + ( c - '0' );
}
struct edge {
int e , next , sign ;
} ed[M] ;
struct Bridge {
int s ,e ;
} bridge[M] ;
int n , m ;
int head[N] ,num ;
int dfn[N] , low[N] ,st[N] ,inst[N] , belong[N] ;
int dp ,top ,cnt ;
int bridgenum ;
int isBridge[N] ;
int faa[N] ;
void add(int s ,int e) {
ed[num].e = e ;
ed[num].sign = 0 ;
ed[num].next = head[s] ;
head[s] = num ++ ;
}
void init() {
mem(dfn, 0) ;
mem(low , 0) ;
mem(st ,0) ;
mem(belong ,0) ;
mem(head, -1) ;
num = 0 ;
dp = 0 ;
top = 0 ;
cnt = 0 ;
bridgenum = 0 ;
mem(isBridge , 0) ;
}
void tarjan(int now ,int fa) {
dfn[now] = low[now] = ++ dp ;
st[top ++ ] = now ;
inst[now] = 1 ;
faa[now] = fa ;
for (int i = head[now] ; ~i ; i = ed[i].next ) {
if(ed[i].sign)continue ;
ed[i].sign = ed[i ^ 1].sign = 1 ;
int e = ed[i].e ;
if(!dfn[e]) {
tarjan(e , now) ;
low[now] = min(low[now] ,low[e]) ;
if(dfn[now] < low[e]) {
bridge[bridgenum].s = now ;
bridge[bridgenum ++ ].e = e ;
isBridge[e] = 1 ;
}
} else if(inst[e]) {
low[now] = min(low[now] , dfn[e]) ;
}
}
if(low[now] == dfn[now]) {
int xx ;
cnt ++ ;
do {
xx = st[-- top] ;
inst[xx] = 0 ;
belong[xx] = cnt ;
} while(xx != now) ;
}
}
void read() {
init() ;
while(m -- ) {
int a , b ;
RD(a) ;
RD(b) ;
add(a , b) ;
add(b , a) ;
}
for (int i = 1 ; i <= n ; i ++ ) {
dp = 0 ,top = 0 ;
if(!dfn[i])tarjan(i , -1) ;
}
}
void LCA(int u ,int v) {
if(dfn[u] < dfn[v]) swap(u , v) ;
while(dfn[u] > dfn[v]) {
if(isBridge[u]) -- bridgenum , isBridge[u] = 0 ;
u = faa[u] ;
}
while(u != v) {
if(isBridge[u])-- bridgenum ,isBridge[u] = 0 ;
u = faa[u] ;
if(isBridge[v])-- bridgenum ,isBridge[v] = 0 ;
v = faa[v] ;
}
}
void solve() {
read() ;
int Q ;
RD(Q) ;
while( Q -- ) {
int a , b ;
RD(a) ;
RD(b) ;
if(belong[a] != belong[b])
LCA(a , b) ;
printf("%d\n",bridgenum) ;
}
}
int main() {
int cas = 0 ;
while(scanf("%d%d",&n ,&m) ,(n + m )) {
printf("Case %d:\n",++ cas) ;
solve() ;
puts("") ;
}
return 0 ;
}
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 2505
#define inf 1<<28
#define LL(x) ( x << 1 )
#define RR(x) ( x << 1 | 1 )
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define PII pair<int,int>
using namespace std;
#define M 500005
#define N 200005
#define stop system("pause")
inline void RD(int &ret) {
char c;
do {
c = getchar();
} while(c < '0' || c > '9') ;
ret = c - '0';
while((c=getchar()) >= '0' && c <= '9')
ret = ret * 10 + ( c - '0' );
}
struct edge {
int e , next , sign ;
} ed[M] ;
struct Bridge {
int s ,e ;
} bridge[M] ;
int n , m ;
int head[N] ,num ;
int dfn[N] , low[N] ,st[N] ,inst[N] , belong[N] ;
int dp ,top ,cnt ;
int bridgenum ;
int isBridge[N] ;
int faa[N] ;
void add(int s ,int e) {
ed[num].e = e ;
ed[num].sign = 0 ;
ed[num].next = head[s] ;
head[s] = num ++ ;
}
void init() {
mem(dfn, 0) ;
mem(low , 0) ;
mem(st ,0) ;
mem(belong ,0) ;
mem(head, -1) ;
num = 0 ;
dp = 0 ;
top = 0 ;
cnt = 0 ;
bridgenum = 0 ;
mem(isBridge , 0) ;
}
void tarjan(int now ,int fa) {
dfn[now] = low[now] = ++ dp ;
st[top ++ ] = now ;
inst[now] = 1 ;
faa[now] = fa ;
for (int i = head[now] ; ~i ; i = ed[i].next ) {
if(ed[i].sign)continue ;
ed[i].sign = ed[i ^ 1].sign = 1 ;
int e = ed[i].e ;
if(!dfn[e]) {
tarjan(e , now) ;
low[now] = min(low[now] ,low[e]) ;
if(dfn[now] < low[e]) {
bridge[bridgenum].s = now ;
bridge[bridgenum ++ ].e = e ;
isBridge[e] = 1 ;
}
} else if(inst[e]) {
low[now] = min(low[now] , dfn[e]) ;
}
}
if(low[now] == dfn[now]) {
int xx ;
cnt ++ ;
do {
xx = st[-- top] ;
inst[xx] = 0 ;
belong[xx] = cnt ;
} while(xx != now) ;
}
}
void read() {
init() ;
while(m -- ) {
int a , b ;
RD(a) ;
RD(b) ;
add(a , b) ;
add(b , a) ;
}
for (int i = 1 ; i <= n ; i ++ ) {
dp = 0 ,top = 0 ;
if(!dfn[i])tarjan(i , -1) ;
}
}
void LCA(int u ,int v) {
if(dfn[u] < dfn[v]) swap(u , v) ;
while(dfn[u] > dfn[v]) {
if(isBridge[u]) -- bridgenum , isBridge[u] = 0 ;
u = faa[u] ;
}
while(u != v) {
if(isBridge[u])-- bridgenum ,isBridge[u] = 0 ;
u = faa[u] ;
if(isBridge[v])-- bridgenum ,isBridge[v] = 0 ;
v = faa[v] ;
}
}
void solve() {
read() ;
int Q ;
RD(Q) ;
while( Q -- ) {
int a , b ;
RD(a) ;
RD(b) ;
if(belong[a] != belong[b])
LCA(a , b) ;
printf("%d\n",bridgenum) ;
}
}
int main() {
int cas = 0 ;
while(scanf("%d%d",&n ,&m) ,(n + m )) {
printf("Case %d:\n",++ cas) ;
solve() ;
puts("") ;
}
return 0 ;
}
