[HDU1599]find the mincost route

Description

[HDU1599]find the mincost route

Solution

恶补图论，最小环问题的板子题
(floyd)来判最小环，复杂度(O(n^3))
枚举(k)从(1)到(n)
最小环的(i)从(1)到(k-1)，(j)从(1)到(i-1)
令(ans=min(ans,f[i][j]+a[i][k]+a[k][j]))
同时维护最短路数组的(i)从(1)到(n)，(j)从(1)到(n)

Code

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x7ffffff
#define MAXN 110
ll n, m, x, y, z, ans;
ll a[MAXN][MAXN], f[MAXN][MAXN];
inline ll read() {
	ll s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w; 
}
int main() {
	while (scanf("%lld%lld", &n, &m) == 2) {
		for (register ll i = 1; i <= n; i++)
			for (register ll j = 1; j <= n; j++)
				a[i][j] = f[i][j] = INF;
		for (register ll i = 1; i <= m; i++) {
			x = read(), y = read(), z = read();
			a[x][y] = a[y][x] = min(a[x][y], z);
			f[x][y] = f[y][x] = min(f[x][y], z);
		}
		ans = INF;
		for (register ll k = 1; k <= n; k++) {
			for (register ll i = 1; i < k; i++)
				for (register ll j = i + 1; j < k; j++)
					ans = min(ans, f[i][j] + a[i][k] + a[k][j]);
			for (register ll i = 1; i <= n; i++)
				for (register ll j = 1; j <= n; j++)
					f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
		}
		if (ans == INF)
			printf("It's impossible.
");
		else
			printf("%lld
", ans);
	}
	return 0;
}