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;
}

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

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原文地址：http://blog.csdn.net/u013923947/article/details/45845735