P1613 跑路 倍增思想 + 邻接矩阵

题意

给定一个有向图，每条边的花费为1。现在有一个空间跑路器，可以走$2^k$长度的路，只用花1秒的时间。问从1走到n最少的时间。
n <= 50, k <= 64。

思路

这道题说是倍增，但是写起来不见倍增的影子，我觉得真妙，对倍增有了更膜拜的认识。
我们可以开一个bool矩阵$dp[i][j][k]$，表示i到j是否可以通过$2^k$的路程到达。更新这个矩阵可以通过类似floyd最短路的思想

if(dp[i][t][k-1] && dp[t][j][k-1]) dp[i][j][k] = 1;

再开一个$dis[i][j]$记录距离，跑一遍floyd就可以了
复杂度是$O（n*n*n*64)$

 

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>


using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/

const int maxn = 109;
int n, m;
int dp[maxn][maxn][maxn];
ll dis[maxn][maxn];

int main() {
    scanf("%d%d", &n, &m);
    rep(i, 1, n) rep(j, 1, n) dis[i][j] = inff;
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        dis[x][y] = 1;
        dp[x][y][0] = 1;
    }

    for (int k = 1; k <= 64; k++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                for (int t = 1; t <= n; t++) {
                    if (dp[i][t][k - 1] && dp[t][j][k - 1]) {
                        dp[i][j][k] = 1;
                        dis[i][j] = 1;
                    }
                }
            }
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= n; k++) {
                if (dis[i][j] > dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
            }
        }
    }

    printf("%lld\n", dis[1][n]);
    return 0;
}