追债之旅

思路

最短路问题，考虑$Dijkstra$，用一个二维$dis[i][j]$数组，表示第$i$天到达$j$号点的最小花费，$dis$数组的更新方式改为$if(dis[day][to]>dis[day-1][now]+value[to]+cost[day])$则更新$dis$数组，所以我们最后只要遍历$i$天到达$n$号节点，也就是$dis[i][n]$数组，最后取其最小值就行。

$Dijkstra$的关键就是一个有能够记录$day,value,pos$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$Dijkstra$板子了。

代码

/*Author : lifehappy
*/
#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 | 1using 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;
}