JZOJ Day9 B组 T2 游历的路线

题目大意：

有$n$个点，每个点之间的边权是一个周期。
求从$1$出发经过$m$条边到$n$的最小权值

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解题思路：

$Solution1$
$dp$
设$f_{i,j}$表示第$i$天到达第$j$个点的最少花费
则得出转移方程:$$f_{i,j}=min(f_{i,j},f_{i-1,k}+a_{k,j,(a_{i,j,0}-1)modk+1)}$$
$a_{i,j,0}$表示周期长度
$a_{i,j,k}$表示从点$i$到点$j$在周期中第$k$天的花费，若$a_{i,j,k}=0$，则周期中的第$k$天不能走

$Solution2$
分层图SPFA
这里不再赘述

$Acceptedcode:$

#include
#include
#include
#define rr register

using namespace std;

const int inf = 1e9;

int n, m;
int a[101][101][21], f[201][101];

inline signed read() {
	int f = 0; char c = getchar();
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) f = f * 10 + c - 48, c = getchar();
	return f;
}

void write(int x) {
	if (x > 9) write(x/10); putchar(x%10+48); return;
}

signed main() {
	/*freopen("lines.in", "r", stdin);
	freopen("lines.out", "w", stdout);*/
	n = read(); m = read();
	for (rr int i = 1; i <= n; ++i) {
		for (rr int j = 1; j <= n; ++j)
			if (i != j) {
				a[i][j][0] = read();
				for (rr int T = 1; T <= a[i][j][0]; ++T)
					a[i][j][T] = read(),
					a[i][j][T] = a[i][j][T] == 0 ? inf : a[i][j][T];
			}
	}
	for (rr int i = 0; i <= m; ++i)
		for (rr int j = 0; j <= n; ++j)
			f[i][j] = inf;
	f[0][1] = 0;
	for (rr int k = 1; k <= m; ++k)
		for (rr int i = 1; i <= n; ++i)
			for (rr int j = 1; j <= n; ++j)
				if (i != j)
					f[k][j] = min(f[k][j], f[k-1][i] + a[i][j][(k-1)%a[i][j][0]+1]);
	write(f[m][n] == inf ? 0 : f[m][n]);
	return 0;
}