UVA - 11280 Flying to Fredericton (伪最短路)

题意：求无向图从起点到终点最多停留k次的最短路

设d[i][j]表示走了i步后到达点j的最小代价，看似最短路，实则dp，因为求解过程中i是递增的，不存在环，直接递推求解即可

什么？你说最短路也属于dp？那没事了

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=100+10,inf=0x3f3f3f3f;
 5 int n,m,hd[N],d[N][N],ne,ka,Q;
 6 char s1[30],s2[30];
 7 struct E {int v,c,nxt;} e[20010];
 8 void link(int u,int v,int c) {e[ne]= {v,c,hd[u]},hd[u]=ne++;}
 9 map<string,int> s2id;
10 int id(string s) {
11     if(s2id.count(s))return s2id[s];
12     int n=s2id.size()+1;
13     return s2id[s]=n;
14 }
15 void solve() {
16     for(int i=0; i<=n; ++i)for(int j=1; j<=n; ++j)d[i][j]=inf;
17     d[0][id("Calgary")]=0;
18     for(int t=0; t<n; ++t)
19         for(int u=1; u<=n; ++u)if(d[t][u]<inf)
20                 for(int i=hd[u]; ~i; i=e[i].nxt) {
21                     int v=e[i].v,c=e[i].c;
22                     d[t+1][v]=min(d[t+1][v],d[t][u]+c);
23                 }
24 }
25 int qry(int k) {
26     int ret=inf;
27     for(int t=0; t<=k; ++t)ret=min(ret,d[t][id("Fredericton")]);
28     return ret;
29 }
30 
31 int main() {
32     int _;
33     for(scanf("%d",&_); _--;) {
34         if(ka++)puts("");
35         printf("Scenario #%d\n",ka);
36         s2id.clear();
37         memset(hd,-1,sizeof hd),ne=0;
38         scanf("%d",&n);
39         for(int i=1; i<=n; ++i)scanf("%*s");
40         scanf("%d",&m);
41         while(m--) {
42             int c;
43             scanf("%s%s%d",s1,s2,&c);
44             link(id(s1),id(s2),c);
45         }
46         solve();
47         for(scanf("%d",&Q); Q--;) {
48             int k;
49             scanf("%d",&k);
50             k++;
51             k=min(k,n);
52             int ans=qry(k);
53             if(ans==inf)puts("No satisfactory flights");
54             else printf("Total cost of flight(s) is $%d\n",ans);
55         }
56     }
57     return 0;
58 }

标签：Fredericton,int,短路,Flying,ne,long,s2id,UVA,hd
来源： https://www.cnblogs.com/asdfsag/p/14618297.html