C. Journey
time limit per test
memory limit per test
input
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
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
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 Ttime 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 ui, vi, ti (1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ ti ≤ 109), meaning that there is a road starting from showplace ui and leading to showplace vi, and Irina spends ti
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
If there are multiple answers, print any of them.
Examples
input
4 3 13
1 2 5
2 3 7
2 4 8
output
3
1 2 4
input
6 6 7
1 2 2
1 3 3
3 6 3
2 4 2
4 6 2
6 5 1
output
4
1 2 4 6
input
5 5 6
1 3 3
3 5 3
1 2 2
2 4 3
4 5 2
output
3
1 3 5
用dp[i][j]表示从1走到i点经过j个点需要的最少时间，在宽搜中进行状态转移
#include <bits/stdc++.h>
#define maxn 5005
#define INF 1000000009
typedef long long ll;
using namespace std;
struct Node{
Node(){
}
Node(int a, int b){
p = a;
t = b;
}
int p, t;
};
vector<Node> v[maxn];
int dp[maxn][maxn], n, m, t;
int pre[maxn][maxn];
queue<Node> q;
void bfs(){
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++){
dp[i][j] = 2e9;
}
dp[1][1] = 0;
q.push(Node(1, 1));
while(!q.empty()){
Node e = q.front();
q.pop();
for(int i = 0; i < v[e.p].size(); i++){
Node h = v[e.p][i];
if(h.t <= t - dp[e.p][e.t]){
if(dp[e.p][e.t] + h.t < dp[h.p][e.t+1]){
dp[h.p][e.t+1] = dp[e.p][e.t] + h.t;
pre[h.p][e.t+1] = e.p * 100000 + e.t;
q.push(Node(h.p, e.t+1));
}
}
}
}
}
void print(int p1, int p2){
if(pre[p1][p2] == 0){
printf("%d", p1);
return ;
}
print(pre[p1][p2] / 100000, pre[p1][p2] % 100000);
printf(" %d", p1);
}
int main(){
// freopen("in.txt", "r", stdin);
int a, b, c;
scanf("%d%d%d", &n, &m, &t);
for(int i = 0; i < m; i++){
scanf("%d%d%d", &a, &b, &c);
v[a].push_back(Node(b, c));
}
bfs();
int ans;
for(int i = n; i >= 1; i--)
if(dp[n][i] <= t){
ans = i;
break;
}
printf("%d\n", ans);
print(n, ans);
puts("");
return 0;
}