POJ 1236 Network of Schools（强连通 Tarjan+缩点）

ACM

题目地址：POJ 1236

题意： 
给定一张有向图，问最少选择几个点能遍历全图，以及最少添加几条边使得有向图成为一个强连通图。

分析： 
跟HDU 2767 Proving Equivalences（题解）一样的题目，不过多了个问题，其实转化成DAG后就不难考虑了，其实只要选择入度为0的点就行了。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        1236.cpp*  Create Date: 2014-07-30 15:13:12*  Descripton:  Tarjan */#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#include <vector>#include <stack>#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 105;vector<int> G[N];stack<int> S;int low[N], dfn[N], sccno[N], tclock, scccnt;int id[N], od[N];int n, rd;void tarjan(int u) {low[u] = dfn[u] = ++tclock;S.push(u);int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (!dfn[v]) {tarjan(v);low[u] = min(low[u], low[v]);} else if (!sccno[v]) {low[u] = min(low[u], dfn[v]);}}if (low[u] == dfn[u]) {scccnt++;int v = -1;while (v != u) {v = S.top();S.pop();sccno[v] = scccnt;}}}void read() {repf (i, 1, n) {G[i].clear();while (scanf("%d", &rd) && rd) {G[i].push_back(rd);}}}void find_scc() {tclock = scccnt = 0;memset(dfn, 0, sizeof(dfn));memset(low, 0, sizeof(low));memset(sccno, 0, sizeof(sccno));repf (i, 1, n) {if (!dfn[i]) {tarjan(i);}}}void solve() {if (scccnt == 1) {printf("1\n0\n");return;}memset(id, 0, sizeof(id));memset(od, 0, sizeof(od));repf (u, 1, n) {int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (sccno[u] != sccno[v]) {id[sccno[v]]++;od[sccno[u]]++;}}}int idnum = 0, odnum = 0;repf (i, 1, scccnt) {idnum += (id[i] == 0);odnum += (od[i] == 0);}printf("%d\n%d\n", idnum, max(idnum, odnum));}int main() {while (~scanf("%d", &n)) {read();find_scc();solve();}return 0;}