Light OJ 1429 - Assassin`s Creed (II)

来源：互联网发布：造价师做题软件编辑：程序博客网时间：2024/04/29 06:13

与 POJ 2594 类似，只不过这里给的是一个有环的图！还有每个人可以走重复的点，任意访问多次都行，只要可达即可。

解法也类似，只不过这里先要缩点，因为同一个强连通分量内的点都相互可达嘛，所以直接缩点就好了！

代码：

#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 DFS(int now,int u){    vis[u] = 1;    for(int i = 0;i < G2[u].size();i ++){        int v = G2[u][i];        if(!vis[v]) {            G[now].push_back(v);            DFS(now, 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 hungary(){    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 T, ca = 0;    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++){            memset(vis, 0, sizeof(vis));            DFS(i,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", ++ca, scc_cnt - hungary());    }    return 0;}

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