题目大意:裸的二分匹配。
题目分析:数据比较强,用来测模版的。这题用hungry跑着会比较吃力,所以用hopcroft-karp算法。这个算法较hungry高效是因为每次bfs找到一个增广路集,然后用dfs进行多路增广,同时找多条增广路,从而效率大增。其实怎么看hk算法都是个没有边权的dinic啊。
参照着wikipedia 敲了一个hk,效率貌似不高啊。。。
详情请见代码:
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 50001;
const int M = 150001;
const int inf = 0x3f3f3f3f; int head[N];
struct node
{
int to,next;
}g[M];
int m,n,p,num;
int matchx[N],matchy[N],que[N],dis[N];
void build(int s,int e)
{
g[num].to = e;
g[num].next = head[s];
head[s] = num ++;
}
bool bfs()
{
int i,j;
int front,rear;
front = rear = 0;
for(i = 1;i <= n;i ++)
{
if(!matchx[i])
{
dis[i] = 0;
que[rear ++] = i;
}
else
dis[i] = inf;
}
dis[0] = inf;
while(front != rear)
{
int u = que[front ++];
if(front == N)
front = 0;
for(i = head[u];~i;i = g[i].next)
{
int v = g[i].to;
if(dis[matchy[v]] == inf)
{
dis[matchy[v]] = dis[u] + 1;
que[rear ++] = matchy[v];
if(rear == N)
rear = 0;
}
}
}
return dis[0] != inf;
}
bool dfs(int u)
{
int i,v;
for(i = head[u];~i;i = g[i].next)
{
v = g[i].to;
if(dis[matchy[v]] == dis[u] + 1)
if(matchy[v] == 0 || dfs(matchy[v]))
{
matchx[u] = v;
matchy[v] = u;
return true;
}
}
dis[u] = inf;
return false;
} void Hopcroft_Karp()
{
memset(matchx,0,sizeof(matchx));
memset(matchy,0,sizeof(matchy));
int ans = 0;
while(bfs())
{
for(int i = 1;i <= n;i ++)
if(!matchx[i])
if(dfs(i))
ans ++;
}
printf("%d\n",ans);
}
int main()
{
int a,b;
while(scanf("%d",&n) != EOF)
{
memset(head,-1,sizeof(head));
num = 1;
scanf("%d%d",&m,&p);
while(p --)
{
scanf("%d%d",&a,&b);
build(a,b);
}
Hopcroft_Karp();
}
return 0;
}