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;
}