匈牙利算法和hopcroft_karp算法模板

//hopcroft_karp算法，复杂度O(sqrt(n)*m)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;

const int N = 320;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, next;
}g[N*N];
int match
, head
;
bool used
;
int p, n;
int nx, ny, cnt, dis; //nx,ny分别是左点集和右点集的点数
int dx
, dy
, cx
, cy
; //dx,dy分别维护左点集和右点集的标号
//cx表示左点集中的点匹配的右点集中的点，cy正好相反
void add_edge(int v, int u)
{
g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
bool bfs() //寻找增广路径集，每次只寻找当前最短的增广路
{
queue<int> que;
dis = INF;
memset(dx, -1, sizeof dx);
memset(dy, -1, sizeof dy);
for(int i = 1; i <= nx; i++)
if(cx[i] == -1) //将未遍历的节点入队，并初始化次节点距离为0
que.push(i), dx[i] = 0;
while(! que.empty())
{
int v = que.front(); que.pop();
if(dx[v] > dis) break;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(dy[u] == -1)
{
dy[u] = dx[v] + 1;
if(cy[u] == -1) dis = dy[u]; //找到了一条增广路，dis为增广路终点的标号
else
dx[cy[u]] = dy[u] + 1, que.push(cy[u]);
}
}
}
return dis != INF;
}
int dfs(int v)
{
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(! used[u] && dy[u] == dx[v] + 1) //如果该点没有被遍历过并且距离为上一节点+1
{
used[u] = true;
if(cy[u] != -1 && dy[u] == dis) continue; //u已被匹配且已到所有存在的增广路终点的标号，再递归寻找也必无增广路，直接跳过
if(cy[u] == -1 || dfs(cy[u]))
{
cy[u] = v, cx[v] = u;
return 1;
}
}
}
return 0;
}
int hopcroft_karp()
{
int res = 0;
memset(cx, -1, sizeof cx);
memset(cy, -1, sizeof cy);
while(bfs())
{
memset(used, 0, sizeof used);
for(int i = 1; i <= nx; i++)
if(cx[i] == -1)
res += dfs(i);
}
return res;
}
int main()
{
int t, a, b;
scanf("%d", &t);
while(t--)
{
cnt = 0;
memset(head, -1, sizeof head);
scanf("%d%d", &p, &n);
for(int i = 1; i <= p; i++)
{
scanf("%d", &a);
for(int j = 0; j < a; j++)
{
scanf("%d", &b);
add_edge(i, b);
}
}
nx = p, ny = n;
if(hopcroft_karp() == p) printf("YES\n");
else printf("NO\n");
}
return 0;
}

//匈牙利算法，复杂度O(nm)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;

const int N = 320;
struct edge
{
int to, next;
}g[N*N];
int match
, head
;
bool use
;
int p, n;
int nx, ny, cnt;
void add_edge(int v, int u)
{
g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
bool dfs(int v)
{
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(use[u] == false)
{
use[u] = true;
if(match[u] == -1 || dfs(match[u]))
{
match[u] = v;
return true;
}
}
}
return false;
}
int hungary()
{
int res = 0;
memset(match, -1, sizeof match);
for(int i = 1; i <= nx; i++)
{
memset(use, 0, sizeof use);
if(dfs(i)) res++;
}
return res;
}
int main()
{
int t, a, b;
scanf("%d", &t);
while(t--)
{
cnt = 0;
memset(head, -1, sizeof head);
scanf("%d%d", &p, &n);
for(int i = 1; i <= p; i++)
{
scanf("%d", &a);
for(int j = 0; j < a; j++)
{
scanf("%d", &b);
add_edge(i, b);
}
}
nx = p, ny = n;
if(hungary() == p) printf("YES\n");
else printf("NO\n");
}
return 0;
}