Codeforces Round #374 (Div. 2) -- C. Journey （DP）

大大体题意：

给你一个有向图，要求从1点出发，到n 点结束，要求在T时间内 经过尽可能多的城市，输出最大数量，和路径！

思路：

dp思想：比赛时自己想的太简单了，果然wa了！

令dp[i][j]表示用i个点  到目的地 j 的最小时间！

那么 直接枚举每个i 点了！

枚举目的地不好转移，那么 就枚举每个边，反正每个边也经过所有的点，也可以完成目的！

那么如果u 能到v！

那么dp[i][v] = min(dp[i-1][u] + w)  时间比dp[i][v]小就可以转移，然后更新上一个节点记录路径！

最后只需要找到  最后一个  dp[i][n] <=T的即可！  这样保证了路径最长 且 时间符合要求！！！

DP递推形式代码：

#include <bits/stdc++.h>using namespace std;const int maxn = 5000 + 10;int dp[maxn][maxn],la[maxn][maxn];const int inf = 0x3f3f3f3f;stack<int>sk;struct Node{int u,v,w;void read(){scanf("%d %d %d",&u, &v, &w);}}p[maxn];int main(){int n,m, k;scanf("%d %d %d",&n, &m, &k);for (int i = 0; i < m; ++i){p[i].read();}memset(dp,inf,sizeof dp);memset(la,-1,sizeof la);dp[1][1] = 0;int pos;for (int i = 2; i <= n; ++i){for (int j = 0; j < m; ++j){int u = p[j].u;int v = p[j].v;int w = p[j].w;if (dp[i-1][u] + w < dp[i][v]){dp[i][v] = dp[i-1][u] + w;la[i][v] = u;//printf("%d %d\n",u,v);//printf("^^^^ %d\n",dp[3][4]);}}if (dp[i][n] <= k){pos = i;}}printf("%d\n",pos);int cur = n;while(cur != -1){sk.push(cur);cur = la[pos--][cur];}//printf("&& %d\n",la[pos][2]);int cnt = 0;while(!sk.empty()){if (cnt++)printf(" ");printf("%d",sk.top());sk.pop();}puts("");return 0;}

DP记忆化搜索代码：（需要仔细处理好边界不合法的情况，RE了几回！）

# include <bits/stdc++.h>using namespace std;const int maxn = 5000 + 10;vector<pair<int,int> >g[maxn];stack<int>sk;struct Node{    int u,v,w;    void read(){        scanf("%d %d %d",&u,&v,&w);    }}p[maxn];int dp[maxn][maxn];int la[maxn][maxn];int dfs(int k,int m){    int& ans = dp[k][m];    if (ans != -1)return ans;    if (k == 1 && m == 1) return ans = 0;    if (k < 1 || m < 1)return 0x3f3f3f3f;    ans = 0x3f3f3f3f;    for (int i = 0; i < g[m].size(); ++i){        int v = g[m][i].first;        int w = g[m][i].second;        int t = dfs(k-1,v) + w;        if (ans > t){            ans = t;            la[k][m] = v;        }    }    return ans;}int main(){    int n,m,T;    scanf("%d %d %d", &n, &m, &T);    for (int i = 0; i < m; ++i){        p[i].read();        g[p[i].v].push_back(make_pair(p[i].u,p[i].w));    }    memset(dp,-1,sizeof dp);    memset(la,-1,sizeof la);    int pos;    for (int i = n; i >= 2; --i){        if (dfs(i,n) <= T){            pos = i;            break;        }    }    printf("%d\n",pos);    int cur = n;    while(cur != -1){        sk.push(cur);        cur = la[pos--][cur];    }    int cnt =0;    while(!sk.empty()){        if (cnt++)printf(" ");        printf("%d",sk.top());        sk.pop();    }    puts("");    return 0;}

C. 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 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 exceedingT. 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 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.

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

4 3 131 2 52 3 72 4 8

output

31 2 4 

input

6 6 71 2 21 3 33 6 32 4 24 6 26 5 1

output

41 2 4 6 

input

5 5 61 3 33 5 31 2 22 4 34 5 2

output

31 3 5