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题目

P1613 跑路

算法标签: 倍增, 动态规划, 最短路

思路

题目给定$2^k$步内都可以$1$秒到达, 因此可以定义一个状态$f[k][i][j]$表示是否存在一条长度为$2^k$的路径使得$i$能走到$j$, 观察数据范围, 点的数量是$50$, 可以直接枚举三个点, 计算出$f$数组, 如果能到达将两个点直接距离标记为$1$, 然后求最短路即可

代码

#include <iostream>
#include <algorithm>
#include <cstring>using namespace std;typedef unsigned long long ULL;
const int N = 60, M = 1e4 + 10, K = 65;int n, m;
int f[K][N][N], d[N][N];int main() {ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);cin >> n >> m;memset(d, 0x3f, sizeof d);for (int i = 1; i <= n; ++i) d[i][i] = 0;for (int i = 0; i < m; ++i) {int u, v;cin >> u >> v;d[u][v] = 1;f[0][u][v] = 1;}for (int k = 1; k < K; ++k) {for (int i = 1; i <= n; ++i) {for (int t = 1; t <= n; ++t) {for (int j = 1; j <= n; ++j) {if (f[k - 1][i][t] && f[k - 1][t][j]) {f[k][i][j] = 1;d[i][j] = min(d[i][j], 1);}}}}}for (int k = 1; k <= n; ++k) {for (int i = 1; i <= n; ++i) {for (int j = 1; j <= n; ++j) {d[i][j] = min(d[i][j], d[i][k] + d[k][j]);}}}int ans = d[1][n];cout << ans << "\n";return 0;
}