codeforces 781 D. Axel and Marston in Bitland （DP+bitset）

D. Axel and Marston in Bitland

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A couple of friends, Axel and Marston are travelling across the country of Bitland. There are n towns in Bitland, with some pairs of towns connected by one-directional roads. Each road in Bitland is either a pedestrian road or a bike road. There can be multiple roads between any pair of towns, and may even be a road from a town to itself. However, no pair of roads shares the starting and the destination towns along with their types simultaneously.

The friends are now located in the town 1 and are planning the travel route. Axel enjoys walking, while Marston prefers biking. In order to choose a route diverse and equally interesting for both friends, they have agreed upon the following procedure for choosing the road types during the travel:

• The route starts with a pedestrian route.
• Suppose that a beginning of the route is written in a string s of letters P (pedestrain road) and B (biking road). Then, the string  is appended to s, where  stands for the string s with each character changed to opposite (that is, all pedestrian roads changed to bike roads, and vice versa).

In the first few steps the route will look as follows: P, PB, PBBP, PBBPBPPB, PBBPBPPBBPPBPBBP, and so on.

After that the friends start travelling from the town 1 via Bitlandian roads, choosing the next road according to the next character of their route type each time. If it is impossible to choose the next road, the friends terminate their travel and fly home instead.

Help the friends to find the longest possible route that can be travelled along roads of Bitland according to the road types choosing procedure described above. If there is such a route with more than 1018 roads in it, print -1 instead.

Input

The first line contains two integers n and m (1 ≤ n ≤ 500, 0 ≤ m ≤ 2n2) — the number of towns and roads in Bitland respectively.

Next m lines describe the roads. i-th of these lines contains three integers vi, ui and ti (1 ≤ vi, ui ≤ n, 0 ≤ ti ≤ 1), where vi and uidenote start and destination towns indices of the i-th road, and ti decribes the type of i-th road (0 for a pedestrian road, 1 for a bike road). It is guaranteed that for each pair of distinct indices i, j such that 1 ≤ i, j ≤ m, either vi ≠ vj, or ui ≠ uj, or ti ≠ tj holds.

Output

If it is possible to find a route with length strictly greater than 1018, print -1. Otherwise, print the maximum length of a suitable path.

Examples

input

2 21 2 02 2 1

output

3

input

2 31 2 02 2 12 2 0

output

-1

Note

In the first sample we can obtain a route of length 3 by travelling along the road 1 from town 1 to town 2, and then following the road 2 twice from town 2 to itself.

In the second sample we can obtain an arbitrarily long route by travelling the road 1 first, and then choosing road 2 or 3 depending on the necessary type.

题目大意：给出一个有向图，边分成两类0/1.每两个点之间不会存在重复的同类边。要求以一条0边开始走，要求每次走的边满足都是前面的取反。简单说就是0->01->0110->01101001->0110100110010110....

题解：DP+bitset

一道不错的DP题。。刚开始被拉过来做，结果想了半天被告知tag读错了。。。。。

f[0/1][i][x][y] 0表示已0打头的串，1表示已1打头的串。i表示长度是2^i，从x出发是否可以到达y

那么转移的时候

if (f[p][i-1][x][y]) f[p][i][x]|=f[p^1][i-1][y]

判断的时候如果f[0][60][1][i]=1，那么答案就是-1

如果不行的话，我们需要求构造一个最长的长度，倒序枚举长度，再枚举上次可到达的点，那么当前可以达到的位置就是now|=f[p][i][x] if (pre[x]) ,每个长度只要有可以到达的点就计入答案。需要注意的是p每到达一个长度都要取反。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#include<bitset>#define N 533#define LL long long using namespace std;bitset<N> f[2][63][N],pre,now;int n,m;LL ans;int main(){freopen("a.in","r",stdin);scanf("%d%d",&n,&m);for (int i=1;i<=m;i++) {int x,y,opt; scanf("%d%d%d",&x,&y,&opt);f[opt][0][x][y]=1;}for (int i=1;i<=60;i++) for (int p=0;p<=1;p++)  for (int x=1;x<=n;x++)    for (int y=1;y<=n;y++)     if (f[p][i-1][x][y]) f[p][i][x]|=f[p^1][i-1][y];for (int i=1;i<=n;i++) if (f[0][60][1][i]) { printf("-1"); return 0; }int p=0; pre[1]=1;for (LL i=59;i>=0;i--) {    now=0;    for (int j=1;j<=n;j++)     if (pre[j]) now|=f[p][i][j];    if (now.count()) p^=1,ans+=(1LL<<i),pre=now;}if (ans>(1e18)) printf("-1\n");else printf("%I64d\n",ans);}