追债之旅：最短路

https://ac.nowcoder.com/acm/problem/14700

思路：这题不是单纯的求1-n的最短路，而是求1-n种路径花费和+挥霍费和最小，所有我们修改dis[n]位dis[n][k],表示花费k天到到1-n的最小花费，同时标记数组为vis[n][k]。同样跑Dijktsra即可。

#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
const int inf = 0x3f3f3f3f;
const int mod = 10007;
const int N = 1e3+7;
const int ds = 1e8+7;
const double PI = 3.141592653589793238462643383;
 
using namespace std;

struct node{
	int d,day,val;
	bool operator < (const node &x)const{
	    return val > x.val;
	}
};

struct edge{
	int v,next,val;
}edge[N*10*5];
int n,m,k;
int head[N],a[N],vis[N][55],cnt = 1;
int dis[N][55];
void add(int u,int v,int w)
{
	edge[cnt].v = v;
	edge[cnt].val = w;
	edge[cnt].next = head[u];
	head[u] = cnt++;
}

void init(){
	for(int i = 0; i <= n; i++) head[i] = -1;
}

void dij(){
//	for(int i = 0; i <= n; i++) vis[i] = 0;
	for(int i = 0; i <= n; i++) 
	    for(int j = 0; j <= k; j++)
			dis[i][j] = ds,vis[i][j] = 0;
	priority_queue<node>q;
	q.push({1,0,0});
	dis[1][0] = 0;
	while(!q.empty()){
		node t = q.top();
		q.pop();
		if(vis[t.d][t.day]) continue;
        //if(t.d == n || t.day >= k) continue;
		vis[t.d][t.day] = 1;
		for(int i = head[t.d]; ~i; i = edge[i].next){
			int v = edge[i].v;
			if(t.day+1 > k) continue;
			if(vis[v][t.day+1]) continue;
			if(dis[v][t.day+1] > dis[t.d][t.day]+a[t.day+1]+edge[i].val){
				dis[v][t.day+1] = dis[t.d][t.day]+a[t.day+1]+edge[i].val;
				q.push({v,t.day+1,dis[v][t.day+1]});
			}
		}
	}
}

void solve() {
	int u,v,w;
	cin >> n >> m >> k;
	init();
	for(int i = 1; i <= m; i++){
		cin >> u >> v >> w;
		add(u,v,w);
		add(v,u,w);
	}
	for(int i = 1; i <= k; i++){
		cin >> a[i];
	}
	dij();
	int ans = ds;
	for(int i = 0; i <= k; i++){
		ans = min(ans,dis[n][i]);
	}
	ans == ds ? cout << "-1" : cout << ans;
}

int main() { 
//	int t;
//	scanf("%d",&t);
//	while(t--)
		solve();
	return 0;
}

标签：const,val,之旅,int,每日,追债,edge,day,dis
来源： https://blog.csdn.net/qq_46653910/article/details/115217551