poj 2449 Remmarguts' Date 第K短路

题意:求T到S的第K短路

做法：

　　SPFA+A*

　　我们将各个点到终点的距离+这个点已走过的距离为启发式函数

　　在终点第K次入队时，此距离为第k短路。

坑点:

　　1.起点终点一样，一开始不算做到达，只有出去再回来才算做到达，因为这个WA了好久。

　　2.用SPFA求各个点到终点的最短距离时，没有将图上的边反向，导致MLE

代码：

 1 //第k短路，自写，有向图，A*+SPFA 
 2 #include<iostream>
 3 using namespace std;
 4 #include<cstring>
 5 #include<vector>
 6 #include<queue>
 7 #include<cstdio>
 8 #define MAXN 1200
 9 int n,m,s,t,k;
10 int d[MAXN];
11 int cs[MAXN];
12 struct teamdata{
13     int dist;
14     int point; 
15     bool operator<(const teamdata&b)const{
16         return this->dist+d[this->point]>b.dist+d[b.point];
17     }
18 }; 
19 struct bian{
20     int val;
21     int point;
22 };
23 vector<bian> maps[MAXN];
24 vector<bian> maps2[MAXN];
25 int zd[MAXN];
26 void SPFA(int s){
27     queue<teamdata> team;
28     while(!team.empty()) team.pop();
29     for (int i=0;i<=n;i++) zd[i]=0x7ffffff;
30     teamdata pt;pt.dist=0;pt.point=s;
31     zd[s]=0;
32     team.push(pt);teamdata pd;
33     while(!team.empty()){
34         pt=team.front();team.pop();
35         int point=pt.point;int distp=pt.dist;
36         for (int i=0;i<maps2[point].size();i++){
37             int nd=maps2[point][i].point;
38             if (zd[nd]>(distp+maps2[point][i].val)){
39                 zd[nd]=distp+maps2[point][i].val;
40                 pd.dist=zd[nd];pd.point=nd;
41                 team.push(pd);
42             }
43         }
44     }
45     for (int i=0;i<=n;i++) d[i]=zd[i];
46 }
47 int short_k(int s,int t,int k){
48     priority_queue<teamdata> priteam;
49     memset(cs,0,sizeof(cs));
50     while(!priteam.empty()) priteam.pop();
51     teamdata pt,pd;
52     pt.dist=0;pt.point=s;priteam.push(pt);
53     int ans=-1;
54     if (d[s]==-1) return -1;
55     while(!priteam.empty()){
56         pt=priteam.top();priteam.pop();
57         cs[pt.point]++;
58     //    cout<<pt.point<<" "<<cs[pt.point]<<" "<<pt.dist<<endl;
59         if (pt.point==t&&cs[pt.point]==k){
60             ans=pt.dist;break;
61         }
62         for (int i=0;i<maps[pt.point].size();i++){
63             pd.dist=pt.dist+maps[pt.point][i].val;
64             pd.point=maps[pt.point][i].point;
65             priteam.push(pd);
66         }
67     }
68     return ans; 
69 }
70 int main(){
71     while(scanf("%d%d",&n,&m)!=EOF){
72         for (int i=0;i<=1010;i++) d[i]=0x7fffffff,maps[i].clear(),maps2[i].clear();
73         bian pt;
74         for (int i=0;i<m;i++){
75             int p,q;int w;
76             scanf("%d%d%d",&p,&q,&w);
77             pt.point=q;pt.val=w;
78             maps[p].push_back(pt);
79             pt.point=p;pt.val=w;
80             maps2[q].push_back(pt);
81         }
82         scanf("%d%d%d",&s,&t,&k);
83         if(s==t) k++; 
84         SPFA(t);
85     //    cout<<"OK"<<endl;
86         int ans=short_k(s,t,k);
87         printf("%d\n",ans);
88     }
89     return 0;
90 }

 

posted @ 2017-10-02 11:01  晓风微微  阅读(196)  评论(0)  编辑  收藏  举报