牛客网NOIP赛前集训营-提高组（第七场）C 洞穴-优快云博客

本文介绍了一种解决洞穴问题的算法，利用矩阵快速幂优化计算过程，重点在于如何仅计算所需的矩阵行，以及预处理矩阵的幂次方。通过实际代码展示了算法的具体实现。

洞穴

思路:

矩阵乘法

由于只需要知道 A ^ l 的第a行第b个元素

所以我们每次在做矩阵乘法时只需要算第a行就可以了

还要像矩阵快速幂一样预处理A ^ (1<<d)

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pli, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int N = 105;
bool a[N][N][35];
bool t[N][2];
int main() {
    int n, m, u, v, q, l;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &u, &v);
        a[u][v][0] = true;
    }
    for (int d = 1; d <= 30; d++) {
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if(a[i][j][d-1])
                for (int k = 1; k <= n; k++) {
                    if(a[j][k][d-1]) {
                        a[i][k][d] = true;
                    }
                }
            }
        }
    }
    scanf("%d", &q);
    while(q--) {
        scanf("%d %d %d", &l, &u, &v);
        int now = 1;
        for (int i = 1; i <= n; i++) t[i][now] = 0;
        t[u][now] = 1;
        for (int i = 30; i >= 0; i--) {
            if(l&(1<<i)) {
                now ^= 1;
                for (int j = 1; j <= n; j++) t[j][now] = 0;
                for (int j = 1; j <= n; j++) {
                    if(t[j][now^1]) {
                        for (int k = 1; k <= n; k++) {
                            if(a[j][k][i])
                                t[k][now] = 1;
                        }
                    }
                }
            }
        }
        if(t[v][now]) puts("YES");
        else puts("NO");
    }
    return 0;
}