[ZPG TEST 117] 跑路【构图】

一看懵了，求一条路的长度whose二进制位中1的个数最小？什么鬼。

其实这种n这么小的图论题，应该往Floyd上想了。令f(p, i, j)为从i走长度为2^p长度的路能否到j，若能，则在一张新的图上连一条i到j的边。最后bfs就猴了。

#include <cstdio>
#include <cstring>

const int maxn = 55, maxm = 10005, maxe = 90000;

int n, m, t1, t2;
bool f[35][maxn][maxn];
int head[maxn], to[maxe], next[maxe], lb;
int que[maxn], head_, tail, h, stp[maxn];

inline void ist(int aa, int ss) {
	to[lb] = ss;
	next[lb] = head[aa];
	head[aa] = lb;
	++lb;
}

int main(void) {
	freopen("run.in", "r", stdin);
	freopen("run.out", "w", stdout);
	memset(head, -1, sizeof head);
	memset(next, -1, sizeof next);
	memset(stp, -1, sizeof stp);
	scanf("%d%d", &n, &m);
	while (m--) {
		scanf("%d%d", &t1, &t2);
		f[0][t1][t2] = true;
		ist(t1, t2);
	}
	for (int p = 1; p < 31; ++p) {
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= n; ++j) {
				for (int k = 1; k <= n; ++k) {
					if (f[p][i][j] = f[p - 1][i][k] && f[p - 1][k][j]) {
						ist(i, j);
						break;
					}
				}
			}
		}
	}
	
	que[tail++] = 1;
	stp[1] = 0;
	while (head_ != tail) {
		h = que[head_++];
		if (h == n) {
			break;
		}
		for (int j = head[h]; j != -1; j = next[j]) {
			if (stp[to[j]] == -1) {
				stp[to[j]] = stp[h] + 1;
				que[tail++] = to[j];
			}
		}
	}
	printf("%d
", stp[n]);
	return 0;
}