「POJ3613」Cow Relays
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「POJ3613」Cow Relays
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就一个思想:(N) 遍 ( ext{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
", res[id[s]][id[t]]);
return 0;
}
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poj3613:Cow Relays(倍增优化+矩阵乘法floyd+快速幂)