UVA 11324 The Largest Clique （强连通缩点 + DAG最长路）

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链接： http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?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;}

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