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

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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;}

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