UVA11324 The Largest Clique

思路：強連通分量縮點轉化DAG+DAG上的DP

解法一：記憶化搜尋

#define mem(a,n) memset(a,n,sizeof(a))
#define f(i,a,b) for(int i=a;i<=b;i++)
#define af(i,a,b) for(int i=a,i>=b;i--)

using namespace std;
typedef long long LL;
const int INF = 20010509;
const int maxn = 1e3 + 100;
const int maxm = 5e4 + 100;

stack<int> s;

int dfs_clock, scc_cnt;
int dfn[maxn], low[maxn], sccno[maxn];
int head[maxn], headnew[maxn], cnt = 0;
int n, m;
int val[maxn], dp[maxn];

struct Edge {
    int next, to;
}e[maxm], enew[maxm];

void add(int from, int to, Edge eset[], int head[]) {
    cnt++;
    eset[cnt].next = head[from];
    eset[cnt].to = to;
    head[from] = cnt;
}

void dfs(int u) {
    dfn[u] = low[u] = ++dfs_clock;
    s.push(u);
    for (int i = head[u]; i; i = e[i].next) {
        int v = e[i].to;
        if (!dfn[v]) {
            dfs(v);
            low[u] = min(low[u], low[v]);
        }
        else if (!sccno[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]) {
        scc_cnt++;
        while (1) {
            int x = s.top(); s.pop();
            sccno[x] = scc_cnt;
            val[scc_cnt]++;
            if (x == u) break;
        }
    }
}

void find_scc(int n) {
    while (!s.empty()) s.pop();
    dfs_clock = scc_cnt = 0;
    mem(sccno, 0);
    mem(dfn, 0);
    mem(low, 0);
    for (int i = 1; i <= n; i++) {
        if (!dfn[i]) dfs(i);
    }
}

int Dp(int i) {
  //實際上是記憶化搜尋
    if (dp[i]) return dp[i];
    dp[i] = val[i];
    for (int j = headnew[i]; j; j = enew[j].next) {
        int v = enew[j].to;
        dp[i] = max(dp[i], Dp(v) + val[i]);
    }
    return dp[i];
}

void init() {
    cnt = 0;
    mem(e, 0);
    mem(enew, 0);
    mem(headnew, 0);
    mem(head, 0);
    mem(val, 0);
    mem(dp, 0);
}
int main(){
    int t; cin >> t; while (t--) {
        init();
        cin >> n >> m;
        for (int i = 1; i <= m; i++) {
            int u, v; cin >> u >> v;
            add(u, v, e, head);
        }
        find_scc(n);
        cnt = 0;
        //將SCC都縮成一個點，建立新圖
        for (int u = 1; u <= n; u++) {
            for (int i = head[u]; i; i = e[i].next) {
                int v = e[i].to;
                if (sccno[u] != sccno[v]) {
                    add(sccno[u], sccno[v], enew, headnew);
                }
            }
        }
        int ans = 0;
        for (int i = 1; i <= scc_cnt; i++) {
            ans = max(ans, Dp(i));
        }
        cout << ans << endl;
    }
    return 0;
}
 

解法二：拓撲排序+DP