Light OJ 1429 Assassin's Creed (ii) BFS + contraction + minimum path Overwrite

Source: Light OJ 1429 Assassin's Creed (II)

Question: at least a few people can go through the full graph and repeat the directed graph.

Ideas: if it is a DAG graph and each vertex cannot be repeated, it means that the bare minimum path overwrites the Dag. Currently, not the Dag may have loops, and each vertex may be repeated. For the graph with a ring, the point can be reduced. it is a DAG graph. In addition, we can repeat the same point as poj 2594 and pre-process the connectivity first.

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <stack>using namespace std;const int maxn = 1010;int vis[maxn];int y[maxn];vector <int> G[maxn], G2[maxn], G3[maxn];int n, m;int a[maxn][maxn];int pre[maxn];int low[maxn];int sccno[maxn];int dfs_clock;int scc_cnt;stack <int> S;void dfs(int u){pre[u] = low[u] = ++dfs_clock;S.push(u);for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(!pre[v]){dfs(v);low[u] = min(low[u], low[v]);}else if(!sccno[v])low[u] = min(low[u], pre[v]);}if(pre[u] == low[u]){scc_cnt++;while(1){int x = S.top();S.pop();sccno[x] = scc_cnt;if(x == u)break;}}}void find_scc(){dfs_clock = scc_cnt = 0;memset(sccno, 0, sizeof(sccno));memset(pre, 0, sizeof(pre));for(int i = 1; i <= n; i++)if(!pre[i])dfs(i);}void BFS(int u){queue <int> Q;memset(vis, 0, sizeof(vis));vis[u] = true;Q.push(u);while(!Q.empty()){int x = Q.front(); Q.pop();for(int i = 0; i < G2[x].size(); i++){int v = G2[x][i];if(vis[v])continue;vis[v] = true;G[u].push_back(v);Q.push(v);}}}bool dfs2(int u){    for(int i = 0; i < G3[u].size(); i++)    {        int v = G3[u][i];        if(vis[v])            continue;        vis[v] = true;        if(y[v] == -1 || dfs2(y[v]))        {            y[v] = u;            return true;        }    }    return false;}int match(){    int ans = 0;    memset(y, -1, sizeof(y));    for(int i = 1; i <= scc_cnt; i++)    {memset(vis, 0, sizeof(vis));if(dfs2(i))    ans++;    }    return ans;}int main(){int cas = 1;int T;scanf("%d", &T);while(T--){scanf("%d %d", &n, &m);for(int i = 0; i <= n; i++)G[i].clear(), G2[i].clear(), G3[i].clear();while(m--){int u, v;scanf("%d %d", &u, &v);G2[u].push_back(v);}for(int i = 1; i <= n; i++)BFS(i);find_scc();for(int u = 1; u <= n; u++){for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(sccno[u] != sccno[v])G3[sccno[u]].push_back(sccno[v]);}}printf("Case %d: %d\n", cas++, scc_cnt-match());}    return 0;}