uva 12167(强连通分支)

题意：在一张有向图中最少添加几条边，能使它强连通。

思路：自己想的时候差了一点。。最后还是看了一眼白书秒过了。首先第一步很简单当然是缩点。变成DAG接下来问题就是DAG上最少添几条边使他强连通。其实只要求出max（出度为零的节点数,入度为零的结点数）。注意当原图已经强联通时特判一下。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-01-31 22:26
 * Filename     : uva_12167.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 100000+10;
vector<int> Map[LEN];
int n, m, dclock, scc_cnt, dfn[LEN], low[LEN], sccn[LEN];
stack<int> s;

void sccinit(){
    for(int i=0; i<LEN; i++) Map[i].clear();
    while(!s.empty()) s.pop();
    memset(dfn, 0, sizeof dfn);
    memset(sccn, 0, sizeof sccn);
    dclock = scc_cnt = 0;
}

void dfs(int u){
    low[u] = dfn[u] = ++dclock;
    s.push(u);
    for(int i=0; i<Map[u].size(); i++){
        int v = Map[u][i];
        if(!dfn[v]){
            dfs(v);
            low[u] = min(low[u], low[v]);
        }else if(!sccn[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]){
         scc_cnt ++;
         while(1){
             int x = s.top();s.pop();
             sccn[x] = scc_cnt;
             if(u == x) break;
         }
    }
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int T, a, b;
    scanf("%d", &T);
    while(T--){
        sccinit();
        scanf("%d%d", &n, &m);
        for(int i=0; i<m; i++){
            scanf("%d%d", &a, &b);
            Map[a].PB(b);
        }
        int od[LEN] = {0}, id[LEN] = {0};
        for(int i=1; i<=n; i++)if(!dfn[i])dfs(i);
        for(int i=1; i<=n; i++){
             for(int j=0; j<Map[i].size(); j++){
                if(sccn[i] == sccn[Map[i][j]]) continue;
                od[sccn[i]] = 1;
                id[sccn[Map[i][j]]] = 1;
             }
        }
        a = b = 0;
        for(int i=1; i<=scc_cnt; i++) {
            if(!od[i]) a++;     
            if(!id[i]) b++;
        }
        if(scc_cnt!=1)printf("%d
", max(a, b));
        else printf("0
");
    }
    return 0;
}