http://codeforces.com/contest/721/problem/C Journey(DAG上的dp)_5000个点,5000条边让你求从1到n的路径长度不超过t中经过点数最多的一条-优快云博客

本文介绍了一个旅行问题，即在一个无环图中寻找从起点到终点不超过特定时间限制的最长路径。通过使用DFS记忆化搜索算法，文章详细展示了如何确定最大访问地点数量及相应路径，并附带了完整的代码实现。

Journey

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Irina arrived to one of the most famous cities of Berland — the Berlatov city. There are n showplaces in the city, numbered from 1 to n, and some of them are connected by one-directional roads. The roads in Berlatov are designed in a way such that there are no cyclic routes between showplaces.

Initially Irina stands at the showplace 1, and the endpoint of her journey is the showplace n. Naturally, Irina wants to visit as much showplaces as she can during her journey. However, Irina's stay in Berlatov is limited and she can't be there for more than T time units.

Help Irina determine how many showplaces she may visit during her journey from showplace 1 to showplace n within a time not exceeding T. It is guaranteed that there is at least one route from showplace 1 to showplace n such that Irina will spend no more than T time units passing it.

Input

The first line of the input contains three integers n, m and T (2 ≤ n ≤ 5000, 1 ≤ m ≤ 5000, 1 ≤ T ≤ 10⁹) — the number of showplaces, the number of roads between them and the time of Irina's stay in Berlatov respectively.

The next m lines describes roads in Berlatov. i-th of them contains 3 integers u_i, v_i, t_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i, 1 ≤ t_i ≤ 10⁹), meaning that there is a road starting from showplace u_i and leading to showplace v_i, and Irina spends t_i time units to pass it. It is guaranteed that the roads do not form cyclic routes.

It is guaranteed, that there is at most one road between each pair of showplaces.

Output

Print the single integer k (2 ≤ k ≤ n) — the maximum number of showplaces that Irina can visit during her journey from showplace 1 to showplace n within time not exceeding T, in the first line.

Print k distinct integers in the second line — indices of showplaces that Irina will visit on her route, in the order of encountering them.

If there are multiple answers, print any of them.

Examples

Input

Output

3
1 2 4

Input

Output

4
1 2 4 6

Input

Output

3
1 3 5

【题意】

DAG图，5000个点，5000条边

让你求从1到n的路径长度不超过T中经过点数最多的一条

解题思路：

int dp[5005][5005]; //f[i][j]表示当前在点i，接下来经过j个点（包括自己）到n点的最小距离

int nex[5005][5005]; //nxt[i][j]表示当前在点i，接下来经过j个点（包括自己）到n点的后继

然后我们对每个点都求出其dp[]和nex[]

注意，这个dfs采取记忆化的方式，起点是n，终点为1这是我写的第一道树形 dp ，不是太难，不懂的可以好好琢磨下这道题，挺好的。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=5005;
vector< pair<int,int> > a[maxn];
int dp[maxn][maxn];
int nex[maxn][maxn];
int vis[maxn];
int n,m,t;
void dfs(int x)
{
    if(vis[x]==1) return;
    vis[x]=1;
    vector< pair<int,int> >::iterator it;
    for(it=a[x].begin(); it!=a[x].end(); ++it)
    {
        dfs(it->first);
        for(int i=2;i<=n;i++){
            int dis=dp[it->first][i-1] + it->second;
            if(dis<dp[x][i]){
                dp[x][i]=dis;
                nex[x][i]=it->first;
            }
        }
    }
}
void print()
{
    for(int i=n;;--i){
        if(dp[1][i]<=t){
            printf("%d\n",i);
            int x=1;
            printf("%d",x);
            while(x!=n){
                x=nex[x][i];
                printf(" %d",x);
                i--;
            }
            printf("\n");
            break;
        }
    }
}
int main()
{
    while(~scanf("%d %d %d",&n,&m,&t))
    {
        int x,y,z;
        for(int i=0;i<m;i++){
            scanf("%d %d %d",&x,&y,&z);
            if(x==n || y==1) continue;
            a[x].push_back(make_pair(y,z));
        }
        memset(vis,0,sizeof(vis));
        memset(dp,0x3f,sizeof(dp));
        memset(nex,0,sizeof(nex));
        dp[n][1]=0;
        vis[n]=1;
        nex[1][1]=n;
        dfs(1);
        print();
    }
    return 0;
}