UVA 11324 - The Largest Clique（强连通分量+缩点)
来源：程序员人生 发布时间：2014-09-26 11:54:48 阅读次数：2529次
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;
}
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