[WC2016]挑战NPC

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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 <bits/stdc++.h>
# 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 <int> 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;
}

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