AC代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof a)
#define N 1009
#define M 20009
int h[N];
int e[M],ne[M],idx,a[20],w[M]; 
ll  dist[20][N];//dist[i][j]表示第i天到大j号点的最小花费 
int vis[20][N];
int n,m,k;
//天数，点的位置，花费 
struct node
{
    ll day,p,val; 
    //改变堆里面的排序规则，使得按照w从小到大排序 
    bool operator < (const node &t) const 
    {
        return val>t.val;
    }
};
ll Read()
{
    char ch;
    ll s=0,w=1;
    ch = getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') w = -1; ch = getchar();}
    while(ch>='0'&&ch<='9'){ s = s*10 +ch-'0'; ch = getchar(); }
    return s*w;
}
void add(int x,int y,int z)
{
    e[idx] = y;
    w[idx] = z;
    ne[idx] = h[x];
    h[x] = idx++;
}
void Dijstra()
{
    mem(dist,INF);
    dist[0][1]=0;//
    priority_queue<node> heap;
    heap.push({0,1,0});
    while(heap.size())
    {
       node tmp = heap.top();
       heap.pop();
       int day = tmp.day,p = tmp.p,val= tmp.val;
       //如果在那天到达过点p，那么就继续下一次循环 
       if(vis[day][p]) continue;
       vis[day][p] = 1;
       for(int i=h[p]; i!=-1; i=ne[i])
       {
               if(day+1>k) continue;
                int j = e[i];//与i相连的点j
                //如果第day+1天到大点j的总花费大于第day天达到p的再花一天到达j的总花费
            //那么更新dist数组 
            if(dist[day+1][j]>val+a[day+1]+w[i])
            {
                dist[day+1][j] = a[day+1]+val+w[i];
                heap.push({day+1,j,dist[day+1][j]});
            } 
       }    
    }
}
int main() {
    mem(h,-1);
    cin>>n>>m>>k;
    for(int i=0; i<m; i++)
    {
        int x,y,z;
//        a =Read();
//        b = Read();
//        c = Read();
        cin>>x>>y>>z;
        add(x,y,z);
        add(y,x,z);
    }
    for(int i=1; i<=k; i++)
    {
        cin>>a[i];
    }
    Dijstra();
    ll ans = INF;
    for(int i=1; i<=k; i++)
       ans = min(ans,dist[i][n]);
    if(ans==INF) printf("-1");
    else cout<<ans<<endl;
    return 0;
}

标签：ch,dist,val,之旅,int,追债,heap,NC14700,day
来源： https://www.cnblogs.com/Jin-yt/p/14328555.html