https://www.luogu.org/problem/show?pid=1613
看到2k就能想到倍增。用一个数组avai[i][j][k]表示点i与点j是否存在长2k的路径,则可以递推出avai[i][j][k]=any{avai[i][v][k-1]&avai[v][j][k-1]},初始值avai[i][i][0]=true。如果avai[i][j][k]==true,则在i点与j点加一条长1的路径。最后BFS或者直接Floyd跑一遍最短路径就可以了。
#include <iostream>
#include <vector>
#include <queue>
#include <cstring>
#define maxn 55
using namespace std; int n, m;
int g[maxn][maxn];
bool avai[maxn][maxn][]; int main()
{
ios::sync_with_stdio(false);
memset(g, 0x3f, sizeof(g));
cin >> n >> m;
int u, v;
for (int i = ; i <= m; i++)
{
cin >> u >> v;
avai[u][v][] = g[u][v] = ;
}
for (int i = ; i <= n; i++)
g[i][i] = ; //f(i,j,k)=f(i,v,k-1)&&f(v,j,k-1)
for (int k = ; k <= ; k++)
{
for (int u = ; u <= n; u++)
{
for (int v = ; v <= n; v++)
{
for (int w = ; w <= n; w++)
{
if (avai[u][v][k] |= avai[u][w][k - ] == && avai[w][v][k - ] == )
g[u][v] = ;
}
}
}
} for (int k = ; k <= n; k++)
for (int i = ; i <= n; i++)
for (int j = ; j <= n; j++)
g[i][j] = min(g[i][k] + g[k][j], g[i][j]); cout << g[][n] << endl;
return ;
}