题意:每组钥匙有两个,只能拿一个。有许多门,每扇门有两把锁(开一把即可),过了一个才能去下一个,问最多能开几扇门
思路:二分能打开的门,建边
1.若门上的锁相同,就一定要选对应的钥匙,比如A,建边A' -> A
2.若门上的锁是一组钥匙的两把,比如A,A',任选一把即可,不用管它
3.若门上的锁是A、B,则A'、B'矛盾(这样不能开门),建边A' -> B,B'-> A
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<list>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#include<numeric>
#include<functional>
using namespace std;
typedef long long ll;
const int maxn = 5005;
map<int,int> ma;
struct edge
{
int to,next;
bool cut;
}ed[maxn];
int head[maxn], tot;
int Low[maxn], DFN[maxn], Stack[maxn];
int Index, top;
bool Instack[maxn];
int block,belong[maxn];
int bridge;
int one[maxn],two[maxn];
void init()
{
memset(head, -1, sizeof(head));
memset(DFN, 0, sizeof(DFN));
memset(Instack, false, sizeof(Instack));
block = tot = 0;
Index = top = 0;
bridge = 0;
}
void Tarjan(int u,int pre)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for(int i = head[u];i != -1;i = ed[i].next)
{
v = ed[i].to;
//if(v == pre)continue; //无向无重边 用,只要不往回走
//if(ed[i].id == pre) continue; //无向图重边判定,不走同一条边
if( !DFN[v] )
{
Tarjan(v,u);
//Tarjan(v,ed[i].id); //无向图重边判定
if( Low[u] > Low[v] )Low[u] = Low[v];
if(Low[v] > DFN[u])
{
bridge++;
ed[i].cut = true;
ed[i^1].cut = true;
}
}
else if( Instack[v] && Low[u] > DFN[v] )
Low[u] = DFN[v];
}
if(Low[u] == DFN[u])
{
block++;
do
{
v = Stack[--top];
Instack[v] = false;
belong[v] = block;
}
while( v!=u );
}
}//使无向图双连通,(叶子+1) / 2
void add(int x,int y)
{
ed[tot].to = y;
ed[tot].next = head[x];
ed[tot].cut = 0;
//ed[tot].id = num; //无向图重边判定
head[x] = tot++;
}
void SCC(int n)
{
for(int i = 0; i < n; i++)
if(!DFN[i])
Tarjan(i,-1);
}
int check(int en,int n)
{
init();
for(int i = 1; i <= en; i++)
{
if(one[i] == two[i])
add(ma[one[i]],one[i]);
else if(ma[one[i]] == two[i])
;
else
{
add(ma[one[i]],two[i]);
add(ma[two[i]],one[i]);
}
}
SCC(n);
int flag = 1;
for(int i = 1; i <= n/2; i++)
{
if(belong[one[i]] == belong[two[i]])
{
flag = 0;
break;
}
}
return flag;
}
int main(void)
{
int n,m,a,b;
while(scanf("%d%d",&n,&m)!=EOF && n+m)
{
ma.clear();
for(int i = 1; i <= n; i++)
{
scanf("%d%d",&a,&b);
ma[a] = b;
ma[b] = a;
}
for(int i = 1; i <= m; i++)
{
scanf("%d%d",&a,&b);
one[i] = a;
two[i] = b;
}
int l = 0,r = m,mid;
while(l < r)
{
int mid = (l+r+1) / 2;
if( check(mid,2*n) )
l = mid;
else
r = mid-1;
}
printf("%d\n",l);
}
return 0;
}