loj 1210 (求最少的加边数使得图变成强连通)

题目链接：http://lightoj.com/volume_showproblem.php?problem=1210

思路：首先是缩点染色，然后重建并且统计新图中的每个点的入度和出度，于是答案就是max(入度为0的点的个数， 出度为0的点的个数，这里有一个trick就是如果scc_count == 1，那么应该输出0.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;

const int MAXN = (20000 + 20);
int n, m, NE;
struct Edge {
    int v, next;
} edge[MAXN << 2];

int head[MAXN];
void Insert(int u, int v)
{
    edge[NE].v = v;
    edge[NE].next = head[u];
    head[u] = NE++;
}

int cnt, scc_count;
int low[MAXN], dfn[MAXN], color[MAXN];
bool mark[MAXN];
stack<int >S;

void Tarjan(int u)
{
    low[u] = dfn[u] = ++cnt;
    mark[u] = true;
    S.push(u);
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (dfn[v] == 0) {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if (mark[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (low[u] == dfn[u]) {
        scc_count++;
        int v;
        do {
            v = S.top();
            S.pop();
            mark[v] = false;
            color[v] = scc_count;
        } while (u != v);
    }
}

int Indegree[MAXN], Outdegree[MAXN];

int main()
{
    int _case, t = 1;
    scanf("%d", &_case);
    while (_case--) {
        scanf("%d %d", &n, &m);
        NE = 0;
        memset(head, -1, sizeof(head));
        while (m--) {
            int u, v;
            scanf("%d %d", &u, &v);
            Insert(u, v);
        }
        cnt = scc_count = 0;
        memset(dfn, 0, sizeof(dfn));
        memset(mark, false, sizeof(mark));
        for (int i = 1; i <= n; i++) {
            if (dfn[i] == 0) Tarjan(i);
        }
        memset(Indegree, 0, sizeof(Indegree));
        memset(Outdegree, 0, sizeof(Outdegree));
        for (int u = 1; u <= n; u++) {
            for (int i = head[u]; i != -1; i = edge[i].next) {
                int v = edge[i].v;
                if (color[u] != color[v]) {
                    Indegree[color[v]]++;
                    Outdegree[color[u]]++;
                }
            }
        }
        int ans1 = 0, ans2 = 0;
        for (int i = 1; i <= scc_count; i++) {
            if (Indegree[i] == 0) ans1++;
            if (Outdegree[i] == 0) ans2++;
        }
        printf("Case %d: ", t++);
        if (scc_count == 1) {
            puts("0");
        } else 
            printf("%d
", max(ans1, ans2));
    }
    return 0;
}