CF 721C DAG上的dp

C. Journey
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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 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

题意：给一个DAG，边有时间，问从1到n在给定时间范围内经过的最多的点有多少个，并输出路径。

做法：一开始zz了，直接搜，爆炸，比如每层两点，就能使路径翻倍。后来看到是DAG，想到先拓扑排序，然后从前往后dp，dp[i][j]表示到第i个点，之前经过j个点的最短时间。最后在dp[n][j]里找一个时间不超过T，并且j最大即可。路径可以用一个二维数组last存。然后ll会MLE，加个特判用int就ok了。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;int n, m;ll p;struct edge{    ll len;    int to;};vector<edge>g[5050];int indegree[5050];bool vis[5050];int ord[5050];int dp[5050][5050];int last[5050][5050];int ans[5050];void toposort(){    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(!indegree[j]&&!vis[j])            {                vis[j]=true;                ord[i]=j;                for(int u=0;u<g[j].size();u++){                    int v=g[j][u].to;                    indegree[v]--;                }                break;            }        }    }}int main(){    memset(indegree, 0, sizeof(indegree));    memset(vis, false, sizeof(vis));    scanf("%d%d%I64d", &n, &m, &p);    for(int i=1;i<=m;i++)    {        int x, y;        ll len;        scanf("%d%d%I64d", &x, &y, &len);        edge tmp;        tmp.to=y;        tmp.len=len;        indegree[y]++;        g[x].push_back(tmp);    }    toposort();    //for(int i=1;i<=n;i++)printf("%d%c", ord[i], i==n?'\n':' ');    bool flag=false;    memset(dp, -1, sizeof(dp));    for(int i=1;i<=n;i++)    {        int u=ord[i];        if(u==1){dp[u][1]=0;flag=true;}        if(flag)        {            for(int j=0;j<g[u].size();j++)            {                edge tmp=g[u][j];                for(int k=0;k<n;k++)                {                    if(dp[u][k]==-1)continue;                    if(dp[tmp.to][k+1]==-1){                        ll tmpr=tmp.len+dp[u][k];                        if(tmpr<=p){                            dp[tmp.to][k+1]=dp[u][k]+tmp.len;last[tmp.to][k+1]=u;                        }                    }                    else {                        if(dp[tmp.to][k+1]>dp[u][k]+tmp.len){                            dp[tmp.to][k+1]=dp[u][k]+tmp.len;                            last[tmp.to][k+1]=u;                        }                        //dp[tmp.to][k+1]=min(dp[tmp.to][k+1], dp[u][k]+tmp.len);                    }                }            }        }        if(u==n)        {            for(int j=n;j>=1;j--)            {                if(dp[u][j]!=-1&&dp[u][j]<=p)                {                    printf("%d\n", j);                    ans[j]=u;                    for(int k=j-1;k>=1;k--)                    {                        ans[k]=last[ans[k+1]][k+1];                    }                    for(int k=1;k<=j;k++)printf("%d%c", ans[k], k==j?'\n':' ');                    break;                }            }            break;        }    }    //for(int i=1;i<=n;i++)    //  for(int j=1;j<=n;j++)printf("%d%c", dp[i][j], j==n?'\n':' ');}