hdu 4685 Prince and Princess（二分图匹配 + 强连通）

题目链接：hdu 4685 Prince and Princess

解题思路

K为当前最大匹配数，王子增加M-K个虚拟点用来匹配没有配对成功的公主，公主增加N-K的虚拟点用来匹配没有配对成功的王子，然后从公主建一条反向边到其配对的王子，做一次强连通，属于同一联通分量的王子和公主可以配对。

代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>

using namespace std;
const int maxn = 2005;

/* Strong connected */
stack<int> S;
vector<int> G[maxn];
int dfsclock, cntscc, sccno[maxn], pre[maxn];

int dfs(int u) {
    int lowu = pre[u] = ++dfsclock;
    S.push(u);

    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (!pre[v]) {
            int lowv = dfs(v);
            lowu = min(lowu, lowv);
        } else if (!sccno[v]) 
            lowu = min(lowu, pre[v]);
    }

    if (lowu == pre[u]) {
        cntscc++;
        while (true) {
            int x = S.top();
            S.pop();
            sccno[x] = cntscc;
            if (x == u) break;
        }
    }
    return lowu;
}

void findSCC(int n) {
    dfsclock = cntscc = 0;
    memset(pre, 0, sizeof(pre));
    memset(sccno, 0, sizeof(sccno));
    for (int i = 0; i < n; i++)
        if (!pre[i]) dfs(i);
}

int N, M, K, BW, L[maxn];
bool T[maxn];

bool match(int u) {
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i] - BW;
        if (!T[v]) {
            T[v] = true;
            if (!L[v] || match(L[v])) {
                L[v] = u;
                return true;
            }
        }
    }
    return false;
}

int KM (int n) {
    int ret = 0;
    memset(L, 0, sizeof(L));
    for (int i = 1; i <= n; i++) {
        memset(T, false, sizeof(T));
        if (match(i)) ret++;
    }
    return ret;
}

void init () {
    scanf("%d%d", &N, &M);
    BW = N + M;
    for (int i = 1; i <= BW * 2; i++) G[i].clear();

    int k, x;
    for (int i = 1; i <= N; i++) {
        scanf("%d", &k);
        while (k--) {
            scanf("%d", &x);
            G[i].push_back(x+BW);
        }
    }
    K = KM(N);
    for (int i = 1; i <= M-K; i++) {
        for (int j = 1; j <= M; j++)
            G[N+i].push_back(j+BW);
    }
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= N-K; j++)
            G[i].push_back(M+j+BW);
    }
    int tmp = KM(N+M-K);
    for (int i = 1; i <= N+M-K; i++)
        G[i+BW].push_back(L[i]);
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();
        findSCC(N+M-K);
        printf("Case #%d:\n", kcas);

        vector<int> ans;
        for (int i = 1; i <= N; i++) {
            ans.clear();
            for (int j = 0; j < G[i].size(); j++) {
                int v = G[i][j];
                if (v - BW > M) continue;
                if (sccno[i] == sccno[v])
                    ans.push_back(v-BW);
            }
            sort(ans.begin(), ans.end());
            printf("%lu", ans.size());
            for (int i = 0; i < ans.size(); i++)
                printf(" %d", ans[i]);
            printf("\n");
        }
    }
    return 0;
}