「JSOI2014」强连通图

「JSOI2014」强连通图

传送门
第一问很显然就是最大的强连通分量的大小。
对于第二问，我们先把原图进行缩点，得到 \(\text{DAG}\) 后，统计出入度为零的点的个数和出度为零的点的个数，两者取 \(\max\) 就是答案。
理性证明可以看这里
参考代码：

#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline T max(T a, T b) { return a > b ? a : b; }
template < class T > inline T min(T a, T b) { return a < b ? a : b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 100005, __ = 2000005;

int tot, head[_]; struct Edge { int ver, nxt; } edge[__];
inline void Add_edge(int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }

int n, m, x[__], y[__];
int siz[_], idgr[_], odgr[_];
int col, co[_], num, dfn[_], low[_], top, stk[_];

inline void tarjan(int u) {
    dfn[u] = low[u] = ++num, stk[++top] = u;
    for (rg int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].ver;
        if (!dfn[v])
            tarjan(v), low[u] = min(low[u], low[v]);
        else
            if (!co[v]) low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) { ++col; do ++siz[co[stk[top]] = col]; while (stk[top--] != u); }
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m);
    for (rg int i = 1; i <= m; ++i) read(x[i]), read(y[i]), Add_edge(x[i], y[i]);
    for (rg int i = 1; i <= n; ++i) if (!dfn[i]) tarjan(i);
    for (rg int i = 1; i <= m; ++i) if (co[x[i]] != co[y[i]]) ++odgr[co[x[i]]], ++idgr[co[y[i]]];
    int ans = 0;
    for (rg int i = 1; i <= col; ++i) ans = max(ans, siz[i]);
    printf("%d\n", ans);
    int icnt = 0, ocnt = 0;
    for (rg int i = 1; i <= col; ++i) icnt += idgr[i] == 0, ocnt += odgr[i] == 0;
    printf("%d\n", max(icnt, ocnt));
    return 0;
}

原文地址：https://www.cnblogs.com/zsbzsb/p/12253738.html