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链接 : http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10294
题意 : 告诉你n个等价的命题 和m个关系 比如 (u,v)代表u可以推导出v, 问至少需要补充多少条边。
用强连通缩点成一张DAG。
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <sstream> #include <cstdio> #include <vector> #include <cmath> #include <queue> #include <stack> #include <set> #include <map> #define lson o<<1, l, m #define rson o<<1|1, m+1, r #define PII pair<int, int> #define ALL(x) x.begin(),x.end() #define mem(a) memset(a,0,sizeof(a)) typedef long long ll; const double pi = acos(-1.0); const int MAX = 0x3f3f3f3f; const ll mod = 1000000007ll; const int N = 20005; const int M = 50005; 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; } stack <int> S; int pre[N], low[N], scc[N], dfs_clock, scc_cnt; 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++; for(;;) { int x = S.top(); S.pop(); scc[x] = scc_cnt; if(x == u) break; } } } void find_scc() { dfs_clock = scc_cnt = 0; mem(scc); mem(pre); for(int i = 1; i <= n; i++) { if(pre[i] == 0) { dfs(i); } } } int in[N], out[N]; 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(); for(int i = 1; i <= scc_cnt; i++) { in[i] = out[i] = 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[u] != scc[v]) { in[scc[v]] = out[scc[u]] = 0; } } } int a = 0, b = 0; for(int i = 1; i <= scc_cnt; i++) { if(in[i]) a++; if(out[i]) b++; } int ans = max(a, b); if(scc_cnt == 1) ans = 0; printf("%d\n", ans); } return 0; }
UVALive - 4287 Proving Equivalences (强连通分量)
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原文地址:http://blog.csdn.net/u013923947/article/details/45845943