二分图最大匹配模版 m√(n) 复杂度

周大爷在比赛中搜到的黑科技二分图模版，复杂度为m√(n)：

注意：点的序号要从0开始！

需要把nx，ny都赋值为n(点数)

const int MAXN = 1010;
const int MAXM = 1010*1010;

struct Edge {
    int v;
    int next;
} edge[MAXM];

struct node {
    double x, y;
    double v;
} a[MAXN], b[MAXN];

int nx, ny;
int cnt;
int t;
int dis;


int first[MAXN];
int xlink[MAXN], ylink[MAXN];
/*xlink[i]表示左集合顶点所匹配的右集合顶点序号，ylink[i]表示右集合i顶点匹配到的左集合顶点序号。*/
int dx[MAXN], dy[MAXN];
/*dx[i]表示左集合i顶点的距离编号，dy[i]表示右集合i顶点的距离编号*/
int vis[MAXN]; //寻找增广路的标记数组

void init() {
    cnt = 0;
    memset(first, -1, sizeof(first));
    memset(xlink, -1, sizeof(xlink));
    memset(ylink, -1, sizeof(ylink));
}

void read_graph(int u, int v) {
    edge[cnt].v = v;
    edge[cnt].next = first[u], first[u] = cnt++;
}

int bfs() {
    queue<int> q;
    dis = INF;
    memset(dx, -1, sizeof(dx));
    memset(dy, -1, sizeof(dy));
    for(int i = 0; i < nx; i++) {
        if(xlink[i] == -1) {
            q.push(i);
            dx[i] = 0;
        }
    }
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        if(dx[u] > dis) break;
        for(int e = first[u]; e != -1; e = edge[e].next) {
            int v = edge[e].v;
            if(dy[v] == -1) {
                dy[v] = dx[u] + 1;
                if(ylink[v] == -1) dis = dy[v];
                else {
                    dx[ylink[v]] = dy[v]+1;
                    q.push(ylink[v]);
                }
            }
        }
    }
    return dis != INF;
}

int find(int u) {
    for(int e = first[u]; e != -1; e = edge[e].next) {
        int v = edge[e].v;
        if(!vis[v] && dy[v] == dx[u]+1) {
            vis[v] = 1;
            if(ylink[v] != -1 && dy[v] == dis) continue;
            if(ylink[v] == -1 || find(ylink[v])) {
                xlink[u] = v, ylink[v] = u;
                return 1;
            }
        }
    }
    return 0;
}

int MaxMatch() {
    int ans = 0;
    while(bfs()) {
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < nx; i++) if(xlink[i] == -1) {
                ans += find(i);
            }
    }
    return ans;
}

double dist(const node a, const node b) {
    return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}

调用：

init();
for(int i = 0; i < m; i++) {
    if(l[edgee[i][0]] && edgee[i][1] != s && !l[edgee[i][1]])    read_graph(edgee[i][0],edgee[i][1]);
    if(l[edgee[i][1]] && edgee[i][0] != s && !l[edgee[i][0]])    read_graph(edgee[i][1],edgee[i][0]);
}
nx = n;
ny = n;
int ans = MaxMatch();