追债之旅

思路

最短路问题，考虑 D i j k s t r a Dijkstra Dijkstra，用一个二维 d i s [ i ] [ j ] dis[i][j] dis[i][j]数组，表示第 i i i天到达 j j j号点的最小花费， d i s dis dis数组的更新方式改为 i f ( d i s [ d a y ] [ t o ] > d i s [ d a y − 1 ] [ n o w ] + v a l u e [ t o ] + c o s t [ d a y ] ) if(dis[day][to] > dis[day - 1][now] + value[to] + cost[day]) if(dis[day][to]>dis[day−1][now]+value[to]+cost[day])则更新 d i s dis dis数组，所以我们最后只要遍历 i i i天到达 n n n号节点，也就是 d i s [ i ] [ n ] dis[i][n] dis[i][n]数组，最后取其最小值就行。

D i j k s t r a Dijkstra Dijkstra的关键就是一个有能够记录 d a y , v a l u e , p o s day, value, pos day,value,pos当前天数，这个状态的最小值，当前位置，这样的结构体，然后重载一下小于号运算符就可以跑个 D i j k s t r a Dijkstra Dijkstra板子了。

代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return x * f;
}
 
const int N1 = 1e3 + 10, N2 = 2e4 + 10;
 
int head[N1], to[N2], nex[N2], value[N2], cnt = 1;
int visit[20][N1], dis[20][N1], cost[20], n, m, k;
 
struct Node {
    int day, pos, value;
    Node(int _day = 0, int _pos = 0, int _value = 0) : day(_day), pos(_pos), value(_value) {}
    bool operator < (const Node & t) const {
        return value > t.value;
    }
};
 
void add(int x, int y, int w) {
    to[cnt] = y;
    nex[cnt] = head[x];
    value[cnt] = w;
    head[x] = cnt++;
}
 
void Dijkstra() {
    for(int i = 0; i <= k; i++)
        for(int j = 0; j <= n; j++)
            dis[i][j] = inf;
    priority_queue<Node> q;
    q.push(Node(0, 1, 0));
    dis[0][1] = 0;
    while(!q.empty()) {
        Node temp = q.top();
        q.pop();
        if(visit[temp.day][temp.pos])   continue;
        visit[temp.day][temp.pos];
        int u = temp.pos, day = temp.day, w = temp.value;
        for(int i = head[u]; i; i = nex[i]) {
            if(day + 1 > k) continue;
            if(dis[day + 1][to[i]] > w + value[i] + cost[day + 1]) {
                dis[day + 1][to[i]] = w + value[i] + cost[day + 1];
                q.push(Node(day + 1, to[i], dis[day + 1][to[i]]));
            }
        }
    }
}
 
int main () {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    n = read(), m = read(), k = read();
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read(), w = read();
        add(x, y, w);
        add(y, x, w);
    }
    for(int i = 1; i <= k; i++)
        cost[i] = read();
    Dijkstra();
    int ans = inf;
    for(int i = 1; i <= k; i++)
        ans = min(ans, dis[i][n]);
    printf("%d\n", ans == inf ? -1 : ans);
    return 0;
}