「JSOI2015」最小表示

「JSOI2015」最小表示

传送门

很显然的一个结论：一条边 \(u \to v\) 能够被删去，当且仅当至少存在一条其它的路径从 \(u\) 通向 \(v\) 。

所以我们就建出正反两张图，对每个点开两个 bitset 维护它与其他点的连通性，这个可以通过拓扑排序预处理。

然后就枚举每一条边，拿两个端点的两个 bitset 与一下即可判断出这条边是否可以删去。

参考代码：

#include <cstdio>
#include <bitset>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
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 _ = 3e4 + 5, __ = 1e5 + 5;
int tot, phead[_], rhead[_]; struct Edge { int v, nxt; } edge[__ << 1];
inline void Add_edge(int* head, int u, int v) { edge[++tot] = (Edge) { v, head[u] }, head[u] = tot; }
int n, m, x[__], y[__], pdgr[_], rdgr[_];
bitset < _ > pbs[_], rbs[_];
inline void toposort(int* head, int* dgr, bitset < _ > * bs) {
    static int hd, tl, Q[_];
    hd = tl = 0;
    for (rg int i = 1; i <= n; ++i) if (!dgr[i]) Q[++tl] = i;
    while (hd < tl) {
	    int u = Q[++hd];
	    for (rg int i = head[u]; i; i = edge[i].nxt) {
	        int v = edge[i].v; bs[v] |= bs[u], bs[v][u] = 1;
	        if (!--dgr[v]) Q[++tl] = v;
	    }
    }
}
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(phead, x[i], y[i]), ++pdgr[y[i]];
	    Add_edge(rhead, y[i], x[i]), ++rdgr[x[i]];
    }
    toposort(phead, pdgr, rbs), toposort(rhead, rdgr, pbs);
    int ans = 0;
    for (rg int i = 1; i <= m; ++i) ans += (pbs[x[i]] & rbs[y[i]]).any();
    printf("%d\n", ans);
    return 0;
}