BZOJ2929 [Poi1999]洞穴攀行

我说这题怎么看着觉得怪怪的。。。

PoPoQQQ给出了解答："给定一个有向图，与起点和终点相连的边只能走一次，剩下的边可以走无数次，问起点到终点可以走多少个人"

题目翻译错了233

好了，那不就是裸的网络流了吗？

/**************************************************************
    Problem: 2929
    User: rausen
    Language: C++
    Result: Accepted
    Time:4 ms
    Memory:1792 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
const int N = 205;
const int M = N * N;
const int inf = (int) 1e9;
   
struct edges{
    int next, to, f;
    edges() {}
    edges(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {}
} e[M << 1];
 
int n, m, S, T;
int tot = 1, first[N];
int d[N], q[N];
 
inline int read() {
    int x = 0;
    char ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}
 
inline void Add_Edges(int x, int y, int z) {
    e[++tot] = edges(first[x], y, z), first[x] = tot;
    e[++tot] = edges(first[y], x, 0), first[y] = tot;
}
   
bool bfs() {
    memset(d, 0, sizeof(d));
    q[1] = S, d[S] = 1;
    int l = 1, r = 1, x, y;
    while (l != r + 1) {
        for (x = first[q[l]]; x; x = e[x].next){
            y = e[x].to;
            if (!d[y] && e[x].f)
                q[(++r) %= N] = y, d[y] = d[q[l]] + 1;
        }
        (++l) %= N;
    }
    return d[T];
}
   
int dinic(int p, int limit) {
    if (p == T || !limit) return limit;
    int x, y, tmp, rest = limit;
    for (x = first[p]; x; x = e[x].next) {
        y = e[x].to;
        if (d[y] == d[p] + 1 && e[x].f && rest) { 
            tmp = dinic(y, min(rest, e[x].f));
            rest -= tmp;
            e[x].f -= tmp, e[x ^ 1].f += tmp;
            if (!rest) return limit;
        }
    }
    if (limit == rest) d[p] = 0;
    return limit - rest;
}
   
int Dinic() {
    int res = 0, x;
    while (bfs())
        res += dinic(S, inf);
    return res;
}
 
int main() {
    int i, j, x;
    n = read();
    for (i = 1; i < n; ++i) {
        for (j = 1, m = read(); j <= m; ++j) {
            x = read();
            Add_Edges(i, x, i == 1 || x == n ? 1 : inf);
        }
    }
    S = 1, T = n;
    printf("%d
", Dinic());
    return 0;
}