bzoj4773 负环 倍增+矩阵

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=4773

题解

最小的负环的长度，等价于最小的 \(len\) 使得存在一条从点 \(i\) 到自己存在一条长度 \(\leq len\) 的负权路径。

为了把 \(\leq len\) 转化为 \(=len\)，我们可以给每一个点建立有个边权为 \(0\) 的自环。

所以考虑倍增邻接矩阵，维护两点之间的经过 \(2^i\) 条边的最短路。

倍增的时候判断走了那么多步有没有负环就可以了。

最后结束的时候再判断一次，防止无解。

---

时间复杂度 \(O(n^3\log n)\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

const int N = 300 + 7;
const int INF = 0x3f3f3f3f;

int n, m;

struct Matrix {
    int a[N][N];
    
    inline Matrix() { memset(a, 0x3f, sizeof(a)); }
    inline Matrix(const int &x) {
        memset(a, 0x3f, sizeof(a));
        for (int i = 1; i <= n; ++i) a[i][i] = x;
    }
    
    inline Matrix operator * (const Matrix &b) {
        Matrix c;
        for (int k = 1; k <= n; ++k)
            for (int i = 1; i <= n; ++i)
                for (int j = 1; j <= n; ++j)
                    smin(c.a[i][j], a[i][k] + b.a[k][j]);
        return c;
    }
} A, B[9];

inline void work() {
    B[0] = A, A = Matrix(0);
    for (int i = 1; i < 9; ++i) B[i] = B[i - 1] * B[i - 1];
    int ans = 0;
    for (int i = 8; ~i; --i) {
        Matrix C = A * B[i];
        int mn = INF;
        for (int j = 1; j <= n; ++j) smin(mn, C.a[j][j]);
        if (mn >= 0) A = C, ans += 1 << i;
    }
    A = A * B[0];
    int mn = INF;
    for (int j = 1; j <= n; ++j) smin(mn, A.a[j][j]);
    if (mn >= 0) puts("0");
    else printf("%d\n", ans + 1);
}

inline void init() {
    read(n), read(m);
    int x, y, z;
    A = Matrix(0);
    for (int i = 1; i <= m; ++i) read(x), read(y), read(z), A.a[x][y] = z;
}

int main() {
#ifdef hzhkk
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}

原文地址：https://www.cnblogs.com/hankeke/p/bzoj4773.html