程序员人生 网站导航

UVA 11324 - The Largest Clique(强连通分量+缩点)

栏目:互联网时间:2014-09-26 11:54:48

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,使得集合内任意两点(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找出最大权值路径即可

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; }


------分隔线----------------------------
------分隔线----------------------------

最新技术推荐