CF576D Flights for Regular Customers

https://www.luogu.com.cn/problem/CF576D

不小心点了一下推荐题目，然后这题就会了

首先每条边按照$d_i$从小到大排序，然后维护邻接矩阵，xjb矩阵快速米即可

可以用bitset优化

草，好像没太想清楚

首先如果是直接min转移是没有前途的，没办法优化

考虑矩阵记为是否能够到达，然后加入一条边后可以很容易得到一个点是否能够达到

然后在跑个bfs加上当前边的边权即可

时间复杂度$O(n^{3}mlogn/w)$

#include<bits/stdc++.h>
#define N 205
using namespace std;
struct E {
    int u, v, c;
} e[N << 1];
int cmp(E x, E y) {
    return x.c < y.c;
}
int n, m, d[N];
struct MT {
    bitset<N> a[N];
    void init() {
        for(int i = 1; i <= n; i ++)
            for(int j = 1; j <= n; j ++)   
                a[i][j] = (i == j);
    }
} a, g;
MT mul(MT a, MT b) {
    MT c; 
    for(int i = 1; i <= n; i ++)
        for(int k = 1; k <= n; k ++)
            if(a.a[i][k])
                c.a[i] |= b.a[k];
    return c;
}
MT qpow(MT x, int y) {
    MT ret; ret.init();
    for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ret = mul(ret, x);
    return ret;
}
void print(MT a) {
    for(int i = 1; i <= n; i ++) {
        for(int j = 1; j <= n; j ++) printf("%d ", (int)a.a[i][j]); printf("\n");
    } printf("\n");
}
queue<int> q;
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i ++) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].c);
    sort(e + 1, e + 1 + m, cmp);
    int lst = 0;
    int ans = 2e9;
    a.init();
    for(int i = 1; i <= m; i ++) {
        if(ans < e[i].c) break;
        int t = e[i].c - lst; lst = e[i].c;
     //   print(a), print(qpow(g, t));
        a = mul(a, qpow(g, t));
        g.a[e[i].u][e[i].v] = 1;
    //    print(a), print(g);
        for(int j = 1; j <= n; j ++) if(a.a[1][j]) d[j] = 0, q.push(j); else d[j] = -1;
        while(q.size()) {
            int u = q.front(); q.pop();
            for(int v = 1; v <= n; v ++) if(g.a[u][v] && d[v] == -1) {
                d[v] = d[u] + 1;
                q.push(v);
            }
        }

        if(d[n] != -1) ans = min(ans, d[n] + lst);
    }
    if(ans == (int)2e9) printf("Impossible");
    else printf("%d", ans);
    return 0;
}