抓住czx【最短路】

题目链接

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  首先，做这样的处理，把每个点的时间分割为几个区间，说明在这个区间内的时候，人在这个点内，那么，我们就有这样的选择，如果在这个区间内，或者区间之前抵达，就说明是可以碰见的，如果在这个区间之后抵达，就说明是见不到的了，所以跑最短路，如果在最短路时间抵达这个点的时候，能找到答案的话，更新答案，并记录即可。

#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 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#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 = 1e5 + 7, maxM = 1e6 + 7;
int N, M, S, T, Q, head[maxN], cnt, a[maxN];
struct Eddge
{
    int nex, to; ll val;
    Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), val(c) {}
}edge[maxM];
inline void addEddge(int u, int v, ll w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, w); }
struct P
{
    int id; ll tim;
    P(int a=0, ll b=0):id(a), tim(b) {}
    friend bool operator < (P e1, P e2) { return e1.tim < e2.tim; }
} que[maxN];
struct node
{
    ll L, R;
    node(ll a=0, ll b=0):L(a), R(b) {}
    friend bool operator < (node e1, node e2) { return e1.R < e2.R; }
};
vector<node> vt[maxN];
ll dis[maxN];
struct Pri_node
{
    int id; ll val;
    Pri_node(int a=0, ll b=0):id(a), val(b) {}
    friend bool operator < (Pri_node e1, Pri_node e2) { return e1.val > e2.val; }
} now;
priority_queue<Pri_node> qq;
ll ans = INF;
inline void Dijkstra()
{
    for(int i=1; i<=N; i++) dis[i] = INF;
    dis[S] = 0;
    qq.push(Pri_node(S, 0));
    int u, len, id; ll w, tmp;
    while(!qq.empty())
    {
        now = qq.top(); qq.pop();
        u = now.id;
        if(now.val > dis[u]) continue;
        len = (int)vt[u].size();
        id = (int)(lower_bound(vt[u].begin(), vt[u].end(), node(dis[u], dis[u])) - vt[u].begin());
        if(id < len)
        {
            tmp = max(dis[u], vt[u][id].L);
            ans = min(ans, tmp);
        }
        for(int i=head[u], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to; w = edge[i].val;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                qq.push(Pri_node(v, dis[v]));
            }
        }
    }
}
inline void init()
{
    cnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d%d%d%d", &N, &M, &S, &T);
    init();
    for(int i=1, u, v, w; i<=M; i++)
    {
        scanf("%d%d%d", &u, &v, &w);
        _add(u, v, w);
    }
    scanf("%d", &Q);
    for(int i=1, pos, tim; i<=Q; i++)
    {
        scanf("%d%d", &tim, &pos);
        que[i] = P(pos, tim);
    }
    que[++Q] = P(T, 0);
    sort(que + 1, que + Q + 1);
    for(int i=1, id; i<Q; i++)
    {
        id = que[i].id;
        vt[id].push_back(node(que[i].tim, que[i + 1].tim - 1));
    }
    vt[que[Q].id].push_back(node(que[Q].tim, INF));
    Dijkstra();
    printf("%lld\n", ans);
    return 0;
}