LightOJ - 1429 Assassin`s Creed (II)(二分图)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int MAXNODE = 1010;
const int MAXEDGE = 10010;

struct Edge{
int u, v, next;
Edge() {}
Edge(int u, int v, int next):u(u), v(v), next(next) {}
}E[MAXEDGE];

//head数组表示头指针,pre数组纪录的是时间戳,lowlink纪录当前节点及其后代所能连回的最早的祖先的时间戳,sccno纪录的是节点是属于哪个强连通分量的
//Stack数组模拟栈,num数组纪录的是每个强连通分量中有多少个节点

int head[MAXNODE], pre[MAXNODE], lowlink[MAXNODE], sccno[MAXNODE], Stack[MAXNODE];

//tot纪录的是边的数量,dfs_clock就是时间戳,top指向栈顶,scc_cnt纪录有多少个强连通分量
int tot, n, m, dfs_clock, top, scc_cnt, cas = 1, All;
bool vis[MAXNODE];
int left[MAXNODE];
vector<int> V[MAXNODE];
vector<int> DAG[MAXNODE];
vector<int> G[MAXNODE];

void init() {
scanf("%d%d", &n, &m);

for (int i = 1; i <= n; i++) {
G[i].clear();
DAG[i].clear();
V[i].clear();
}

int u, v;
for (int i = 0; i < m; i++) {
scanf("%d%d", &u, &v);
G[u].push_back(v);
}
}

void dfs(int u) {
pre[u] = lowlink[u] = ++dfs_clock;//纪录当前节点的时间戳
Stack[++top] = u;

for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
//如果下一个节点没有访问过
if (!pre[v]) {
dfs(v);
lowlink[u] = min(lowlink[u], lowlink[v]);
}//如果下一个节点已经被访问过了,且不属于任何一个连通分量
else if (!sccno[v]) lowlink[u] = min(lowlink[u], pre[v]);
}

int x;
//满足强连通分量的要求，当前节点u为该强连通分量最早发现的节点
if (pre[u] == lowlink[u]) {
scc_cnt++;
while (1) {
x = Stack[top--];
sccno[x] = scc_cnt;
if (x == u) break;
}
}
}

bool dfs2(int u) {
for (int i = 0; i < V[u].size(); i++) {
int v = V[u][i];
if (vis[v]) continue;
vis[v] = true;
if (!left[v] || dfs2(left[v])) {
left[v] = u;
return true;
}
}
return false;
}

void bfs(int cur) {
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(cur);
vis[cur] = true;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (vis[v]) continue;
vis[v] = true;
DAG[cur].push_back(v);
Q.push(v);
}
}
}

void solve() {
memset(pre, 0, sizeof(pre));
memset(sccno, 0, sizeof(sccno));
dfs_clock = top = scc_cnt = 0;

for (int i = 1; i <= n; i++)
bfs(i);

for (int i = 1; i <= n; i++)
if (!pre[i]) dfs(i);

int u, v;
for (int i = 1; i <= n; i++) {
for (int j = 0; j < DAG[i].size(); j++) {
u = sccno[i];
v = sccno[DAG[i][j]];
if (u == v) continue;
V[u].push_back(v);
}
}

int ans = 0;
memset(left, 0, sizeof(left));
for (int i = 1; i <= scc_cnt; i++) {
memset(vis, 0, sizeof(vis));
if (dfs2(i)) ans++;
}
printf("Case %d: %d\n", cas++, scc_cnt - ans);
}

int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}