POJ1062 昂贵的聘礼

点权拆出来与0号点连边,按照下属→上级的方向建立有向图,跑一个s=0,t=1的单源最短路

限制条件:搜索到的最短路径中,阶级差不超过m

 1 #include <queue> 
 2 #include <vector>
 3 #include <cstring>
 4 #include <iostream>
 5 using namespace std;
 6 #define N 105
 7 #define INF 1000000
 8 struct edge{
 9     int v,w;
10     edge(int v,int w):v(v),w(w){}
11 };
12 vector<edge>G[N];
13 int level[N],dis[N],vis[N];
14 
15 #define ifIsUnderTheRestrictionOfLevel if(minLevel<=level[v]&&level[v]<=maxLevel)
16 
17 void spfa(int n,int minLevel,int maxLevel){
18     queue<int>q;
19     fill(dis,dis+n+1,INF),fill(vis,vis+n+1,0);
20     q.push(0),vis[0]=1,dis[0]=0;    
21     while(!q.empty()) {
22         int u=q.front();q.pop(),vis[u]=0;
23         for(int i=0;i<G[u].size();i++) {
24             edge&e=G[u][i];
25             int&v=e.v,w=e.w;
26             ifIsUnderTheRestrictionOfLevel
27             if (dis[v]>dis[u]+w) {
28                 dis[v]=dis[u]+w;
29                 if(!vis[v])q.push(v),vis[v]=1;                    
30             }
31         }
32     }
33 }
34 
35 int main(){
36     int m,n;
37     cin>>m>>n;
38     for(int i=1;i<=n;i++){
39         int P,L,X;
40         cin>>P>>L>>X;
41         level[i]=L;
42         G[0].push_back(edge(i,P));
43         while(X--){
44             int T,V;
45             cin>>T>>V;
46             G[T].push_back(edge(i,V));
47         }
48     }        
49     int ans=INF;
50     for(int i=level[1]-m;i<=level[1];i++){
51         spfa(n,i,i+m);
52         ans=min(ans,dis[1]);
53     }
54     cout<<ans<<endl;
55     return 0;
56 }
~~Jason_liu O(∩_∩)O
原文地址:https://www.cnblogs.com/JasonCow/p/14959809.html