Sol

这做法我是想不到\(TAT\)

每个筐子拆成三个相互连边

球向三个筐子连边

然后跑一般图最大匹配

这三个筐子间最多有一个匹配

那么显然每个球一定会放在一个筐子里，一定有一个匹配

如果筐子间有匹配，则有一个半空的筐子，因为它一定只匹配了小于等于\(1\)个球

答案为匹配数\(-n\)

使答案最大即匹配数最大

上带花树就好了

# include 
    
     # define RG register# define IL inline# define Fill(a, b) memset(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(1005);const int __(2e5 + 5);typedef int Arr[_];IL int Input(){    RG int x = 0, z = 1; RG char c = getchar();    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);    return x * z;}Arr first, match, fa, vis, tim, pre;int n, m, cnt, idx, ans, E, t1, t2, t3;queue 
     
       Q;struct Edge{    int to, next;} edge[__];IL void Add(RG int u, RG int v){    edge[cnt] = (Edge){v, first[u]}, first[u] = cnt++;    edge[cnt] = (Edge){u, first[v]}, first[v] = cnt++;}IL int Find(RG int x){    return x == fa[x] ? x : fa[x] = Find(fa[x]);}IL int LCA(RG int x, RG int y){    ++idx, x = Find(x), y = Find(y);    while(tim[x] != idx){        tim[x] = idx;        x = Find(pre[match[x]]);        if(y) swap(x, y);    }    return x;}IL void Blossom(RG int x, RG int y, RG int p){    while(Find(x) != p){        pre[x] = y, y = match[x];        if(vis[y] == 2) vis[y] = 1, Q.push(y);        if(Find(x) == x) fa[x] = p;        if(Find(y) == y) fa[y] = p;        x = pre[y];    }}IL int Aug(RG int S){    for(RG int i = 1; i <= t3; ++i) vis[i] = pre[i] = 0, fa[i] = i;    while(!Q.empty()) Q.pop();    Q.push(S), vis[S] = 1;    while(!Q.empty()){        RG int u = Q.front(); Q.pop();        for(RG int e = first[u]; e != -1; e = edge[e].next){            RG int v = edge[e].to;            if(Find(v) == Find(u) || vis[v] == 2) continue;            if(!vis[v]){                vis[v] = 2, pre[v] = u;                if(!match[v]){                    for(RG int x = v, lst; x; x = lst)                        lst = match[pre[x]], match[pre[x]] = x, match[x] = pre[x];                    return 1;                }                vis[match[v]] = 1, Q.push(match[v]);            }            else{                RG int p = LCA(u, v);                Blossom(u, v, p);                Blossom(v, u, p);            }        }    }    return 0;}int main(RG int argc, RG char *argv[]){    for(RG int T = Input(); T; --T){        n = Input(), m = Input(), E = Input();        t1 = n + m, t2 = t1 + m, t3 = t2 + m;        ans = cnt = idx = 0;        for(RG int i = 1; i <= t3; ++i) first[i] = -1, match[i] = 0, tim[i] = 0;        for(RG int i = 1; i <= m; ++i)            Add(n + i, t1 + i), Add(t1 + i, t2 + i), Add(n + i, t2 + i);        for(RG int i = 1, u, v; i <= E; ++i){            u = Input(), v = Input();            Add(u, n + v), Add(u, t1 + v), Add(u, t2 + v);        }        for(RG int i = 1; i <= t3; ++i) if(!match[i]) ans += Aug(i);        printf("%d\n", ans - n);    }    return 0;}