HDU 2767 Proving Equivalences（强连通 Tarjan+缩点）

ACM

题目地址：HDU 2767

题意： 
给定一张有向图，问最少添加几条边使得有向图成为一个强连通图。

分析： 
Tarjan入门经典题，用tarjan缩点，然后就变成一个有向无环图(DAG)了。 
我们要考虑的问题是让它变成强连通，让DAG变成强连通就是把尾和头连起来，也就是入度和出度为0的点。 
统计DAG入度和出度，然后计算头尾，最大的那个就是所求。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        2767.cpp*  Create Date: 2014-07-30 14:30:55*  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 = 20010;stack<int> S;vector<int> G[N];int dfn[N], low[N], sccno[N], tclock, scccnt;int id[N], od[N];int t, n, m, x, y;void tarjan(int u) {dfn[u] = low[u] = ++tclock;S.push(u);int sz = G[u].size();repf (i, 0, sz - 1) {int v = G[u][i];if (!dfn[v]) {// v not visitedtarjan(v);low[u] = min(low[u], low[v]);} else if (!sccno[v]) {// v not belong to scc, so it was in the stacklow[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 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 read() {scanf("%d%d", &n, &m);repf (i, 0, n)G[i].clear();while (m--) {scanf("%d%d", &x, &y);G[x].push_back(y);}}int solve() {if (scccnt == 1)return 0;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);}return max(idnum, odnum);}int main() {scanf("%d", &t);while (t--) {read();find_scc();printf("%d\n", solve());}return 0;}