Yaroslav and Time CodeForces - 302D

原题链接

考察:最短路

思路:

        就是最短路....由贪心知道去曾今去过的点只会平白无故损失能量,所以要尽量不走旧点,可以跑一遍dijkstra

        g[n][n]需要long long,不知道是不是因为这个反复WA....

 1 #include <iostream>
 2 #include <cstring>
 3 #include <queue>
 4 using namespace std;
 5 typedef long long LL;
 6 typedef pair<int,int> PII;
 7 const int N = 110;
 8 const LL INF = 0x3f3f3f3f3f3f3f3f;
 9 int n,d,g[N][N];
10 LL dist[N];
11 bool st[N];
12 struct Node{
13     int x,y,w;
14 }node[N];
15 LL dijkstra()
16 {
17     for(int i=1;i<=n;i++) dist[i] = (1ll<<63)-1;
18     dist[1] = 0;
19     priority_queue<PII,vector<PII>,greater<PII> > q;
20     q.push({0,1});
21     while(q.size())
22     {
23         PII it = q.top();
24         q.pop();
25         int u = it.second;
26         if(st[u]) continue;
27         st[u] = 1;
28         for(int i=1;i<=n;i++)
29         {
30             if(u!=i&&!st[i]&&dist[i]>dist[u]+g[u][i])
31             {
32                 dist[i] = dist[u]+g[u][i];
33                 q.push({dist[i],i});
34             }
35         }
36     }
37     return dist[n];
38 }
39 int main()
40 {
41     scanf("%d%d",&n,&d);
42     for(int i=2;i<n;i++) scanf("%d",&node[i].w);
43     for(int i=1;i<=n;i++) scanf("%d%d",&node[i].x,&node[i].y);
44     for(int i=1;i<=n;i++)
45       for(int j=1;j<=n;j++)
46         if(i!=j)
47             g[i][j] = d*(abs(node[i].x-node[j].x)+abs(node[i].y-node[j].y))-node[j].w;
48           else g[i][j] = INF;
49     printf("%lld\n",dijkstra());
50     return 0;
51 }

 

posted @ 2021-03-31 01:27  acmloser  阅读(48)  评论(0)  编辑  收藏  举报