拯救小tim【最短路】

题目链接

---

  这里有一个坑点，譬如说，我们从S出发的时间，不是刚好卡着第一个的“起始点”，没准出发的第一步，没有卡起始点，而是在后面的到达其他点的时候卡了起始点这样的情况，所以我们应该从0～max_BegTim的来枚举起点时间，然后跑Dijkstra即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-4
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e2 + 7;
int N, M, S, T;
struct Eddge
{
    int v, ts, td, c;
    Eddge(int a=0, int s=0, int d=0, int b=0):v(a), ts(s), td(d), c(b) {}
};
vector<Eddge> vt[maxN];
int dis[maxN];
struct node
{
    int id, val;
    node(int a=0, int b=0):id(a), val(b) {}
    friend bool operator < (node e1, node e2) { return e1.val > e2.val; }
} now;
priority_queue<node> Q;
inline int Dijkstra(int bt)
{
    while(!Q.empty()) Q.pop();
    for(int i=1; i<=N; i++) dis[i] = INF;
    dis[S] = bt;
    Q.push(node(S, bt));
    int u, len, v, w;
    while(!Q.empty())
    {
        now = Q.top(); Q.pop();
        u = now.id;
        if(dis[u] < now.val) continue;
        len = (int)vt[u].size();
        for(int i=0; i<len; i++)
        {
            v = vt[u][i].v; w = vt[u][i].c;
            if(max(dis[u], vt[u][i].ts) + w > vt[u][i].td) continue;
            if(max(dis[u], vt[u][i].ts) + w < dis[v])
            {
                dis[v] = max(dis[u], vt[u][i].ts) + w;
                Q.push(node(v, dis[v]));
            }
        }
    }
    return dis[T] == INF ? INF : dis[T] - bt;
}
int main()
{
    scanf("%d%d%d%d", &N, &M, &S, &T);
    for(int i=1, u, v, ts, td, w; i<=M; i++)
    {
        scanf("%d%d%d%d%d", &u, &v, &ts, &td, &w);
        if(td - ts < w) continue;
        vt[u].push_back(Eddge(v, ts, td, w));
    }
    int len = (int)vt[S].size(), mxt = 0;
    for(int i=0; i<len; i++) mxt = max(mxt, vt[S][i].ts);
    int ans = INF;
    for(int i=0; i<=mxt; i++)
    {
        ans = min(ans, Dijkstra(i));
    }
    if(ans == INF) printf("Impossible\n");
    else printf("%d\n", ans);
    return 0;
}