3 seconds
256 megabytes
standard input
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 from1 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 nocyclic 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 Ttime units passing it.
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 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.
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.
|
1 2 3 4 |
4 3 13 1 2 5 2 3 7 2 4 8 |
|
1 2 |
3 1 2 4 |
|
1 2 3 4 5 6 7 |
6 6 7 1 2 2 1 3 3 3 6 3 2 4 2 4 6 2 6 5 1 |
|
1 2 |
4 1 2 4 6 |
|
1 2 3 4 5 6 |
5 5 6 1 3 3 3 5 3 1 2 2 2 4 3 4 5 2 |
|
1 2 |
3 1 3 5 |
在比赛的时候想的是去用BFS穷举，最后也不了了之地WA了，后来对着赛后题解敲了一遍DP解法，发觉自己对DP真的是太陌生了，其实这道题目明眼人一看就应该知道用DP。
首先令 \(dp_{i,j}\) 为在已经到达\(j\)个点过后到达\(i\)所需要的最小的时间，由此可得需要求得就是在\(i\)为\(n\)的时候\(j\)的值是多少，\(1\le j \le n\)，但是如果\(dp_{n,n}\)比T大的话就不行，所以令\(j\)初始值为\(n\)，然后逐步减一，直至\(dp_{n,j} \le T\)，这样就找到了最多可以走过的showplaces，最后输出求解dp过程中保存的路径，其中状态转移方程为：
\(dp_{i,j} = \min ((dp_{to_i,j-1} +w_i | to_i \in E_i), dp_{i,j})\)\(E_i\) 为从\(i\)出发的所有边（对于每一条边\((u,v)\)，保存为\((v,u)\))，\(to_i\)为\(E_i\)的到达地。
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 |
/* * File Name: 721C.cpp * Author: razrLeLe * Mail: razrlele@gmail.com */ #include <vector> #include <set> #include <deque> #include <queue> #include <algorithm> #include <functional> #include <iostream> #include <cstdio> #include <cmath> #include <cstdlib> #include <string> #include <cstring> #include <string.h> #include <map> #include <cctype> #include <list> #include <stack> using namespace std; typedef long long ll; #define INF 0x3f3f3f3f #define LOCAL const int maxn = 5007; int dp[maxn][maxn]; int path[maxn][maxn]; vector<int> edge[maxn]; vector<int> cost[maxn]; int n, m; int T; int calc(int i, int j){ if( dp[i][j] != -1){ return dp[i][j]; } else{ dp[i][j] = INF; for(int index = 0; index < edge[i].size(); index++){ int value = calc(edge[i][index], j-1)+cost[i][index]; if( value < dp[i][j]){ dp[i][j] = value; path[i][j] = edge[i][index]; } } return dp[i][j]; } } int main(int argc, char ** argv) { #ifdef LOCAL freopen("/Users/yuyue/build/data.txt", "r", stdin); // freopen("/Users/yuyue/build/out.txt", "w", stdout); #endif cin >> n >> m >> T; memset(dp, -1, sizeof(int)*maxn*maxn); dp[1][1] = 0; int u, v, time; int ans = 0; for(int i= 1; i <= m; i++){ cin >> v >> u >> time; edge[u].push_back(v); cost[u].push_back(time); } for(int i = n; i >= 1; i--){ if(calc(n, i) <= T){ ans = i; break; } } printf("%d\n", ans); vector<int> final_path; final_path.push_back(n); int now = n; while (ans>1){ final_path.push_back(path[now][ans]); now =path[now][ans]; ans--; } for(int i = final_path.size()-1; i > 0; i--) printf("%d ", final_path[i]); printf("%d\n", final_path[0]); return 0; } |