「POJ3613」Cow Relays

「POJ3613」Cow Relays

传送门
就一个思想:\(N\) 遍 \(\text{Floyd}\) 求出经过 \(N\) 个点的最短路
看一眼数据范围,想到离散化+矩阵快速幂
代码:

#include <cstring>
#include <cstdio>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
template < class T > inline void chkmin(T& a, const T& b) { if (a > b) a = b; }
template < class T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while ('0' > c || c > '9') f |= c == '-', c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 502, __ = 1e6 + 5;

int n, m, s, t, tot, id[__];

struct Matrix {
    int a[_][_];
    inline void init() { memset(a, 0x3f, sizeof a); }
    inline int* operator [] (const int& id) { return a[id]; }
    inline Matrix operator * (const Matrix& b) const {
        Matrix ans; ans.init();
        for (rg int k = 1; k <= tot; ++k)
            for (rg int i = 1; i <= tot; ++i)
                for (rg int j = 1; j <= tot; ++j)
                    chkmin(ans.a[i][j], a[i][k] + b.a[k][j]);
        return ans;
    }
} f;

inline Matrix power(Matrix x, int k) {
    Matrix res = x; --k;
    for (; k; k >>= 1, x = x * x)
        if (k & 1) res = res * x;
    return res;
}

int main() {
#ifndef ONLINE_JUDGE
    file("cpp");
#endif
    read(n), read(m), read(s), read(t), f.init();
    for (rg int u, v, w; m--; ) {
        read(w), read(u), read(v);
        if (!id[u]) id[u] = ++tot;
        if (!id[v]) id[v] = ++tot;
        f[id[u]][id[v]] = f[id[v]][id[u]] = w;
    }
    Matrix res = power(f, n);
    printf("%d\n", res[id[s]][id[t]]);
    return 0;
}