BZOJ 5335 智力竞赛

大致题意：给出一个DAG，问能否用n+1条可重复路径覆盖整个图。

最小有重复路径覆盖问题，先传递闭包，转化成无重复路径覆盖问题。
然后把原图每个点拆成两个点建立二分图，然后用原图点数$-$最大匹配数就是答案。
如果可以覆盖就输出$AK$，否则二分一个最大可行权值$mid$，大于$mid$的点连一个$i->i$的边，表示忽略这个点，然后照常建图即可。

#define others
#ifdef poj
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#endif // poj
#ifdef others
#include 
#include 
#include 
#endif // others
//#define file
#define all(x) x.begin(), x.end()
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define eps 1e-8
const double pi = acos(-1.0);


typedef long long LL;
typedef long long DLL;
typedef unsigned long long ULL;
void umax(LL &a, LL b) {
    a = max(a, b);
}
void umin(LL &a, LL b) {
    a = min(a, b);
}
int dcmp(double x) {
    return fabs(x) <= eps?0:(x > 0?1:-1);
}
void file() {
    freopen("data_in.txt", "r", stdin);
    freopen("data_out.txt", "w", stdout);
}

DLL mod = 1e9;

DLL Pow(DLL a,DLL b) {
    DLL res=1;
    a%=mod;
    for(; b; b>>=1) {
        if(b&1)res=res*a%mod;
        a=a*a%mod;
    }
    return res;
}
//
//void print(DLL x) {
//    if(x < 0) {
//        x = -x;
//        putchar('-');
//    }
//    if(x > 9) print(x/10);
//    putchar(x%10 + '0');
//}
//#define iostart
//#define iostart
#define pb(x) push_back(x)
namespace solver {
    const int maxn = 510;
    int n, m;
    int g[maxn][maxn];
    vector<int> G[maxn];
    int link[maxn], vis[maxn];
    int v[maxn];
    bool dfs(int u) {
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i];
            if(!vis[v]) {
                vis[v] = 1;
                if(link[v] == -1 || dfs(link[v])) {
                    link[v] = u;
                    return 1;
                }
            }
        }
        return 0;
    }
    int gao() {
        int ans = 0;
        memset(link, -1, sizeof link);
        for(int i = 1; i <= m; i++) {
            memset(vis, 0, sizeof vis);
            if(dfs(i))
                ans++;
        }
        return ans;
    }
    bool check_val(int x) {
        for (int i = 1; i <= m; i++) {
            G[i].clear();
        }
        for (int i = 1; i <= m; i++){
            if (v[i] > x) {
                G[i].push_back(i);
            }
            for (int j = 1; j <= m; j++){
                if (g[i][j]) {
                    G[i].push_back(j);
//                    cout << i << " xx " << j << endl;
                }
            }
        }
        int match = gao();
//        cout << x << " " << m - match << endl;
        return m - match <= n + 1;
    }
    vector<int> hash_val;
    void solve() {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= m; i++){
            int val, k;
            scanf("%d%d", &val, &k);
            v[i] = val;
            hash_val.push_back(val);
            while (k--) {
                int to;
                scanf("%d", &to);
                g[i][to] = 1;
            }
        }
        for (int k = 1; k <= m; k++) {
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= m; j++) {
                    if (g[i][k] && g[k][j]) {
                        g[i][j] = 1;
                    }
                }
            }
        }
        sort(all(hash_val));
        hash_val.erase(unique(all(hash_val)), hash_val.end());
        int L = 0, R = (int)hash_val.size() - 2, M;
        while (L + 4 < R) {
            M = L + R >> 1;
            if (check_val(hash_val[M])) {
                L = M;
            } else {
                R = M;
            }
        }
        for (; L < R; L++) {
            if (!check_val(hash_val[L]))
                break;
        }
        if (check_val(hash_val[L])) {
            puts("AK");
        } else {
            printf("%d\n", hash_val[L]);
        }
    }
}

int main() {
#ifdef iostart
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif // iostart
//    file();
    solver::solve();
    return 0;
}

由于传递闭包之后是一个稠密图，所以可以用bitset去优化匈牙利算法。

#define others
#ifdef poj
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#endif // poj
#ifdef others
#include 
#include 
#include 
#endif // others
//#define file
#define all(x) x.begin(), x.end()
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
#define eps 1e-8
const double pi = acos(-1.0);


typedef long long LL;
typedef long long DLL;
typedef unsigned long long ULL;
void umax(LL &a, LL b) {
    a = max(a, b);
}
void umin(LL &a, LL b) {
    a = min(a, b);
}
int dcmp(double x) {
    return fabs(x) <= eps?0:(x > 0?1:-1);
}
void file() {
    freopen("data_in.txt", "r", stdin);
    freopen("data_out.txt", "w", stdout);
}

DLL mod = 1e9;

DLL Pow(DLL a,DLL b) {
    DLL res=1;
    a%=mod;
    for(; b; b>>=1) {
        if(b&1)res=res*a%mod;
        a=a*a%mod;
    }
    return res;
}
//
//void print(DLL x) {
//    if(x < 0) {
//        x = -x;
//        putchar('-');
//    }
//    if(x > 9) print(x/10);
//    putchar(x%10 + '0');
//}
//#define iostart
//#define iostart
#define pb(x) push_back(x)
namespace solver {
    const int maxn = 510;
    int n, m;
    int g[maxn][maxn];
    int a[maxn][maxn/32+1], b[maxn][maxn/32+1];
    int link[maxn], vis[maxn];
    int q[maxn];

    typedef int U;
    int tot = m >> 5;
    inline void set1(U v[],int x){v[x>>5]|=1U<<(x&31);}
    inline void flip(U v[],int x){v[x>>5]^=1U<<(x&31);}
    bool dfs(int u) {
        for (int i = 0; i <= tot; i++) {
            for (;;) {
                U o = b[u][i]&vis[i];
                if (!o) break;
                int y = i<<5|__builtin_ctz(o);
                flip(vis, y);
                if (!link[y] || dfs(link[y])) return link[y]=u, 1;
            }
        }
        return 0;
    }
    int gao() {
        for (int i = 1; i <= m; i++) link[i] = 0;
        int ans = 0;
        for(int i = 1; i <= m; i++) {
            for (int j = 1; j <= m; j++) set1(vis, j);
            if(dfs(i)) ans++;
        }
        return ans;
    }
    bool check_val(int x) {
        for (int i = 1; i <= m; i++){
            for (int j = 0; j <= tot; j++) {
                b[i][j] = a[i][j];
            }
        }
        for (int i = 1; i <= m; i++) {
            if (q[i] > x) {
                set1(b[i], i);
            }
        }
        for (int i = 1; i <= m; i++) link[i] = 0;
        int match = gao();
        return m - match <= n + 1;
    }
    vector<int> hash_val;
    void solve() {
        scanf("%d%d", &n, &m);
        tot = m >> 5;
        for (int i = 1; i <= m; i++){
            int val, k;
            scanf("%d%d", &val, &k);
            q[i] = val;
            hash_val.push_back(val);
            while (k--) {
                int to;
                scanf("%d", &to);
                g[i][to] = 1;
            }
        }
        for (int k = 1; k <= m; k++) {
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= m; j++) {
                    if (g[i][k] && g[k][j]) {
                        g[i][j] = 1;
                    }
                }
            }
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= m; j++) {
                if (g[i][j])
                    set1(a[i], j);
            }
        }
        sort(all(hash_val));
        hash_val.erase(unique(all(hash_val)), hash_val.end());
        int L = 0, R = (int)hash_val.size() - 2, M;
        while (L + 4 < R) {
            M = L + R >> 1;
            if (check_val(hash_val[M])) {
                L = M;
            } else {
                R = M;
            }
        }
        for (; L < R; L++) {
            if (!check_val(hash_val[L]))
                break;
        }
        if (check_val(hash_val[L])) {
            puts("AK");
        } else {
            printf("%d\n", hash_val[L]);
        }
    }
}

int main() {
#ifdef iostart
    ios::sync_with_stdio(0);
    cin.tie(0);
#endif // iostart
//    file();
    solver::solve();
    return 0;
}