经(luo)典(ti) K短路题目= =
流程:
先把所有边逆向,做一遍dijkstra,得到估价函数h(x)(x到T的最短路距离)
f(x)=g(x)+h(x)
按f(x)维护一个堆……T第k次出堆时的g(T)即为ans
另外,需要特判:如果S==T,k++
1 Source Code
2 Problem: 2449 User: sdfzyhy
3 Memory: 11260K Time: 141MS
4 Language: G++ Result: Accepted
5
6 Source Code
7
8 //POJ 2449
9 #include<queue>
10 #include<cstdio>
11 #include<cstring>
12 #include<cstdlib>
13 #include<iostream>
14 #include<algorithm>
15 #define rep(i,n) for(int i=0;i<n;++i)
16 #define F(i,j,n) for(int i=j;i<=n;++i)
17 #define D(i,j,n) for(int i=j;i>=n;--i)
18 #define pb push_back
19 using namespace std;
20 typedef long long LL;
21 inline int getint(){
22 int r=1,v=0; char ch=getchar();
23 for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1;
24 for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch;
25 return r*v;
26 }
27 const int N=1010,M=100010,INF=0x3f3f3f3f;
28 /*******************template********************/
29 int to[2][M],next[2][M],head[2][N],len[2][M],cnt[2];
30 void ins(int x,int y,int z,int k){
31 to[k][++cnt[k]]=y; next[k][cnt[k]]=head[k][x]; head[k][x]=cnt[k]; len[k][cnt[k]]=z;
32 }
33 #define f(i,x,k) for(int i=head[k][x],y=to[k][i];i;i=next[k][i],y=to[k][i])
34
35 int n,m,K,S,T;
36 int d[N],times[N],from[N],route[N];
37 bool vis[N];
38 typedef pair<int,int>pii;
39 #define mp make_pair
40 void dij(){
41 priority_queue<pii,vector<pii>,greater<pii> >Q;
42 memset(d,0x3f,sizeof d);
43 d[T]=0;
44 Q.push(mp(0,T));
45 while(!Q.empty()){
46 int x=Q.top().second; Q.pop();
47 if (vis[x]) continue;
48 vis[x]=1;
49 f(i,x,0)
50 if (!vis[y] && d[y]>d[x]+len[0][i]){
51 d[y]=d[x]+len[0][i];
52 Q.push(mp(d[y],y));
53 }
54 }
55 // F(i,1,n) printf("%d ",d[i]); puts("");
56 }
57
58 struct node{
59 LL w,to;
60 bool operator < (const node &b)const {
61 return w+d[to] > b.w+d[b.to];
62 }
63 };
64 LL astar(){
65 priority_queue<node>Q;
66 memset(times,0,sizeof times);
67 if (d[S]==INF) return -1;
68 Q.push((node){0,S});
69 while(!Q.empty()){
70 LL x=Q.top().to,w=Q.top().w; Q.pop();
71 // printf("%lld %lld\n",x,w);
72 times[x]++;
73 if (x==T && times[T]==K) return w;
74 if (times[x]>K) continue;
75 f(i,x,1) Q.push((node){w+len[1][i],y});
76 }
77 return -1;
78 }
79
80 int main(){
81 #ifndef ONLINE_JUDGE
82 freopen("2449.in","r",stdin);
83 freopen("2449.out","w",stdout);
84 #endif
85 n=getint(); m=getint();
86 F(i,1,m){
87 int x=getint(),y=getint(),z=getint();
88 ins(x,y,z,1); ins(y,x,z,0);
89 }
90 S=getint(); T=getint(); K=getint();
91 if (S==T) K++;
92 dij();
93 printf("%lld\n",astar());
94 return 0;
95 }
Time Limit: 4000MS | Memory Limit: 65536K | |
Total Submissions: 23008 | Accepted: 6295 |
Description
"Prince Remmarguts lives in his kingdom UDF – United Delta of Freedom. One day their neighboring country sent them Princess Uyuw on a diplomatic mission."
"Erenow, the princess sent Remmarguts a letter, informing him that she would come to the hall and hold commercial talks with UDF if and only if the prince go and meet her via the K-th shortest path. (in fact, Uyuw does not want to come at all)"
Being interested in the trade development and such a lovely girl, Prince Remmarguts really became enamored. He needs you - the prime minister's help!
DETAILS: UDF's capital consists of N stations. The hall is numbered S, while the station numbered T denotes prince' current place. M muddy directed sideways connect some of the stations. Remmarguts' path to welcome the princess might include the same station twice or more than twice, even it is the station with number S or T. Different paths with same length will be considered disparate.
Input
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).
Output
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14
Source