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  • UVA 11324

    UVA 11324 - The Largest Clique

    题目链接

    题意:给定一个有向图,要求找一个集合,使得集合内随意两点(u, v)要么u能到v,要么v能到u,问最大能选几个点

    思路:强连通分量,构造出scc之后,缩点,每一个点的权值是集合点个数,然后做一遍dag找出最大权值路径就可以

    代码:

    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <stack>
    #include <algorithm>
    using namespace std;
    
    const int N = 20005;
    
    vector<int> g[N], scc[N];
    int pre[N], lowlink[N], sccno[N], dfs_clock, scc_cnt;
    stack<int> S;
    
    void dfs_scc(int u) {
    	pre[u] = lowlink[u] = ++dfs_clock;
    	S.push(u);
    	for (int i = 0; i < g[u].size(); i++) {
    		int v = g[u][i];
    		if (!pre[v]) {
    			dfs_scc(v);
    			lowlink[u] = min(lowlink[u], lowlink[v]);
    		} else if (!sccno[v])
    			lowlink[u] = min(lowlink[u], pre[v]);
    	}
    	if (lowlink[u] == pre[u]) {
    		scc_cnt++;
    		while (1) {
    			int x = S.top(); S.pop();
    			sccno[x] = scc_cnt;
    			if (x == u) break;
    		}
    	}
    }
    
    void find_scc(int n) {
    	dfs_clock = scc_cnt = 0;
    	memset(sccno, 0, sizeof(sccno));
    	memset(pre, 0, sizeof(pre));
    	for (int i = 0; i < n; i++)
    		if (!pre[i]) dfs_scc(i);
    }
    
    int t, n, m, val[N];
    vector<int> g2[N];
    
    void build() {
    	memset(val, 0, sizeof(val));
    	for (int i = 1; i <= scc_cnt; i++)
    		g2[i].clear();
    	for (int u = 0; u < n; u++) {
    		val[sccno[u]]++;
    		for (int j = 0; j < g[u].size(); j++) {
    			int v = g[u][j];
    			if (sccno[u] != sccno[v])
    				g2[sccno[u]].push_back(sccno[v]);
    		}
    	}
    }
    
    int dp[N];
    
    int dfs(int u) {
    	if (dp[u] != -1) return dp[u];
    	dp[u] = val[u];
    	for (int i = 0; i < g2[u].size(); i++) {
    		int v = g2[u][i];
    		dp[u] = max(dp[u], dfs(v) + val[u]);
    	}
    	return dp[u];
    }
    
    int main() {
    	scanf("%d", &t);
    	while (t--) {
    		scanf("%d%d", &n, &m);
    		for (int i = 0; i < n; i++)
    			g[i].clear();
    		int u, v;
    		while (m--) {
    			scanf("%d%d", &u, &v);
    			u--; v--;
    			g[u].push_back(v);
    		}
    		find_scc(n);
    		build();
    		memset(dp, -1, sizeof(dp));
    		int ans = 0;
    		for (int i = 0; i < n; i++)
    			ans = max(ans, dfs(sccno[i]));
    		printf("%d
    ", ans);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/zfyouxi/p/4272786.html
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