• 牛客网NOIP赛前集训营-提高组(第七场)C 洞穴


    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;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9885009.html
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