K短路/A*
经(luo)典(ti) K短路题目= =
K短路学习：http://www.cnblogs.com/Hilda/p/3226692.html
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++
Source Code
Problem: User: sdfzyhy
Memory: 11260K Time: 141MS
Language: G++ Result: Accepted Source Code //POJ 2449
#include<queue>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
#define pb push_back
using namespace std;
typedef long long LL;
inline int getint(){
int r=,v=; char ch=getchar();
for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
return r*v;
}
const int N=,M=,INF=0x3f3f3f3f;
/*******************template********************/
int to[][M],next[][M],head[][N],len[][M],cnt[];
void ins(int x,int y,int z,int k){
to[k][++cnt[k]]=y; next[k][cnt[k]]=head[k][x]; head[k][x]=cnt[k]; len[k][cnt[k]]=z;
}
#define f(i,x,k) for(int i=head[k][x],y=to[k][i];i;i=next[k][i],y=to[k][i]) int n,m,K,S,T;
int d[N],times[N],from[N],route[N];
bool vis[N];
typedef pair<int,int>pii;
#define mp make_pair
void dij(){
priority_queue<pii,vector<pii>,greater<pii> >Q;
memset(d,0x3f,sizeof d);
d[T]=;
Q.push(mp(,T));
while(!Q.empty()){
int x=Q.top().second; Q.pop();
if (vis[x]) continue;
vis[x]=;
f(i,x,)
if (!vis[y] && d[y]>d[x]+len[][i]){
d[y]=d[x]+len[][i];
Q.push(mp(d[y],y));
}
}
// F(i,1,n) printf("%d ",d[i]); puts("");
} struct node{
LL w,to;
bool operator < (const node &b)const {
return w+d[to] > b.w+d[b.to];
}
};
LL astar(){
priority_queue<node>Q;
memset(times,,sizeof times);
if (d[S]==INF) return -;
Q.push((node){,S});
while(!Q.empty()){
LL x=Q.top().to,w=Q.top().w; Q.pop();
// printf("%lld %lld\n",x,w);
times[x]++;
if (x==T && times[T]==K) return w;
if (times[x]>K) continue;
f(i,x,) Q.push((node){w+len[][i],y});
}
return -;
} int main(){
#ifndef ONLINE_JUDGE
freopen("2449.in","r",stdin);
freopen("2449.out","w",stdout);
#endif
n=getint(); m=getint();
F(i,,m){
int x=getint(),y=getint(),z=getint();
ins(x,y,z,); ins(y,x,z,);
}
S=getint(); T=getint(); K=getint();
if (S==T) K++;
dij();
printf("%lld\n",astar());
return ;
}
| 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
first line contains two integer numbers N and M (1 <= N <= 1000, 0
<= M <= 100000). Stations are numbered from 1 to N. Each of the
following M lines contains three integer numbers A, B and T (1 <= A, B
<= N, 1 <= T <= 100). It shows that there is a directed
sideway from A-th station to B-th station with time T.
The last line consists of three integer numbers S, T and K (1 <= S, T <= N, 1 <= K <= 1000).
Output
single line consisting of a single integer number: the length (time
required) to welcome Princess Uyuw using the K-th shortest path. If K-th
shortest path does not exist, you should output "-1" (without quotes)
instead.
Sample Input
2 2
1 2 5
2 1 4
1 2 2
Sample Output
14
Source