链接： ?id=30726
题意：有向图G，求一个最大的点集，使得点集中任意两个节点u和v，满足要么u可以到达v，要么v可以到达u，或者u和v可以相互到达。
可以强连通缩点成一张DAG，以为每个强连通分量要么选要么不选。求DAG上的最长路二次建图用了2种不同的方法，，也分别用了记忆花搜索DP和直接递推DP
vector建图和记忆化搜索：
#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <stack>#include <cmath>#include <map>#define lson o<<1,l,m#define rson o<<1|1,m+1,r#define mem(a) memset(a,0,sizeof(a))typedef long long ll;const int N = 1005;const int M = 50005;const ll mod = 1000000007;using namespace std;int n, m, T;int he[N];struct C {int ne, to;} e[M];void add(int id, int x, int y) {e[id].to = y;e[id].ne = he[x];he[x] = id;}int pre[N], low[N], scc[N], dfs_clock, scc_cnt;stack <int> S;void dfs(int u) {pre[u] = low[u] = ++ dfs_clock;S.push(u);for(int i = he[u]; i != -1; i = e[i].ne) {int v = e[i].to;if(pre[v] == 0) {dfs(v);low[u] = min(low[u], low[v]);} else if(scc[v] == 0) {low[u] = min(low[u], pre[v]);}}if(low[u] == pre[u]) {scc_cnt ++;while(1) {int x = S.top(); S.pop();scc[x] = scc_cnt;if(x == u) break;}}}void find_scc() {mem(scc);mem(pre);dfs_clock = scc_cnt = 0;for(int i = 1; i <= n; i++) {if(pre[i] == 0) dfs(i);}}int dp[N], vis[N], w[N];vector <int> G[N];int DP(int u) {int& ans = dp[u];if(vis[u] == 1) return ans;vis[u] = 1;ans = w[u];for(int i = 0; i < G[u].size(); i++) {int v = G[u][i];ans = max(ans, DP(v) + w[u]);}return ans;}int main() {cin >> T;while(T–) {cin >> n >> m;memset(he, -1, sizeof(he));for(int i = 1; i <= m; i++) {int x, y;scanf("%d%d", &x, &y);add(i, x, y);}find_scc();mem(w);for(int i = 1; i <= n; i++) {int x = scc[i];w[x]++;}for(int i = 1; i <= scc_cnt; i++) {G[i].clear();}int id = 1;for(int u = 1; u <= n; u++) {for(int i = he[u]; i != -1; i = e[i].ne) {int v = e[i].to;if(scc[v] != scc[u]) {G[scc[u]].push_back(scc[v]);}}}memset(vis, 0, sizeof(vis));int ans = 0;for(int i = 1; i <= scc_cnt; i++) {ans = max(ans, DP(i));}printf("%d\n", ans);}return 0;}
下面是数组邻接表建图递推DP（类似最长上升子序列）
#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <stack>#include <cmath>#include <map>#define lson o<<1,l,m#define rson o<<1|1,m+1,r#define mem(a) memset(a,0,sizeof(a))typedef long long ll;const int N = 1005;const int M = 50005;const ll mod = 1000000007;using namespace std;int n, m, T;int he[N];struct C {int ne, to;} e[M];void add(int id, int x, int y) {e[id].to = y;e[id].ne = he[x];he[x] = id;}int pre[N], low[N], scc[N], dfs_clock, scc_cnt;stack <int> S;void dfs(int u) {pre[u] = low[u] = ++ dfs_clock;S.push(u);for(int i = he[u]; i != -1; i = e[i].ne) {int v = e[i].to;if(pre[v] == 0) {dfs(v);low[u] = min(low[u], low[v]);} else if(scc[v] == 0) {low[u] = min(low[u], pre[v]);}}if(low[u] == pre[u]) {scc_cnt ++;while(1) {int x = S.top(); S.pop();scc[x] = scc_cnt;if(x == u) break;}}}void find_scc() {mem(scc);mem(pre);dfs_clock = scc_cnt = 0;for(int i = 1; i <= n; i++) {if(pre[i] == 0) dfs(i);}}int w[N], he1[N];struct C1 {int ne, to;} e1[N];void add1(int id, int x, int y) {e1[id].to = y;e1[id].ne = he1[x];he1[x] = id;}int dp[N], vis[N];int DP(int u) {int& ans = dp[u];if(vis[u] == 1) return ans;vis[u] = 1;ans = w[u];for(int i = he1[u]; i != -1; i = e1[i].ne) {int v = e1[i].to;ans = max(ans, DP(v) + w[v]);}return ans;}int main() {cin >> T;while(T–) {cin >> n >> m;memset(he, -1, sizeof(he));for(int i = 1; i <= m; i++) {int x, y;scanf("%d%d", &x, &y);add(i, x, y);}find_scc();mem(w);for(int i = 1; i <= n; i++) {int x = scc[i];w[x]++;}map < pair<int, int>, int > mp;int id = 1;memset(he1, -1, sizeof(he1));for(int u = 1; u <= n; u++) {for(int i = he[u]; i != -1; i = e[i].ne) {int v = e[i].to;if(scc[v] != scc[u]) {if(mp[ make_pair(scc[u], scc[v]) ] == 0) {mp[ make_pair(scc[u], scc[v]) ] = 1;add1(id, scc[u], scc[v]);id++;}}}}memset(vis, 0, sizeof(vis));int ans = 0;for(int i = 1; i <= scc_cnt; i++) {dp[i] = w[i];for(int j = 1; j < i; j++) {int u;for(u = he1[i]; u != -1; u = e1[u].ne) {if(e1[u].to == j) {break;}}if(u == -1) continue;dp[i] = max(dp[i], dp[j] + w[i]);}ans = max(ans, dp[i]);}printf("%d\n", ans);}return 0;}
最可怕的敌人，就是没有坚强的信念。
