【SCC缩点+DAGdp】P1137 旅行计划

P1137 旅行计划

思路：缩点+DAGdp，与模板有所区别的是，本题求的是“以节点(i)为终点的最长路”。将缩点之后的新图反向建边，即可直接用“以节点(i)为起点的最长路”的DAGdp。

#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 = 1e5 + 100;
const int maxm = 2e5 + 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 dp[maxn];
int val_new[maxn];

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

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

//dp[i]:=以节点i为终点的最长路
//dp[j]=max(dp[i],dp[j])（其中j满足：存在i→j的有向边）

int Dp(int i) {
    if (dp[i]) return dp[i];
    dp[i] = val_new[i];
    for (int j = headnew[i]; j; j = enew[j].next) {
        int v = enew[j].to;
        dp[i] = max(dp[i], Dp(v) + val_new[i]);
    }
    return dp[i];
}

int main(){
    cnt = 0;
    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;
    //将SCC都缩成一个点，建立新图
    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[v], sccno[u], enew, headnew);
               //反向建图，dp时按“dp[i]:=以i为起点的最长路”dp
            }
        }
    }
    
    //以每个城市作为终点
    for (int i = 1; i <= n; i++) {
        cout << Dp(sccno[i]) << endl;
    }
    return 0;
}