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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 10098 | Accepted: 3620 |
Description
Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.
The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.
The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).
Input
Output
Sample Input
4 4 1 2 100 2 4 200 2 3 250 3 4 100
Sample Output
450
Hint
Source
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #define Max 10000 #define inf 1<<28 using namespace std; int S,T,K,n,m; int head[Max],rehead[Max]; int num,renum; int dis[Max]; bool visit[Max]; int ans[Max]; int qe[Max*10]; struct kdq{ int v,len,next; } edge[200005],reedge[200005]; struct a_star { //A*搜索时的优先级队列; int v; int len; bool operator<(const a_star &a)const{ //f(i)=d[i]+g[i] return len+dis[v]>a.len+dis[a.v]; } }; void insert(int u,int v,int len){//正图和逆图 edge[num].v=v; edge[num].len=len; edge[num].next=head[u]; head[u]=num; num++; reedge[renum].v=u; reedge[renum].len=len; reedge[renum].next=rehead[v]; rehead[v]=renum; renum++; } void init(){ memset(ans,0,sizeof(ans)); for(int i=0; i<=n; i++) head[i]=-1,rehead[i]=-1; num=1,renum=1; } int ans1; void spfa(){//从T开始求出T到所有点的 dis[] int i,j; for(i=1; i<=n; i++) dis[i]=inf; dis[T]=0; visit[T]=1; int num=0,cnt=0; qe[num++]=T; while(num>cnt){ int temp=qe[cnt++]; visit[temp]=0; for(i=rehead[temp]; i!=-1 ; i=reedge[i].next){ int tt=reedge[i].v; int ttt=reedge[i].len; if(dis[tt]>dis[temp]+ttt) { dis[tt]=dis[temp]+ttt; if(!visit[tt]) { qe[num++]=tt; visit[tt]=1; } } } } } int A_star(){ if(S==T) K++; if(dis[S]==inf) return -1; a_star n1; n1.v=S; n1.len=0; priority_queue <a_star> q; q.push(n1); while(!q.empty()){ a_star temp=q.top(); q.pop(); ans[temp.v]++; if(ans[T]==K){//当第K次取到T的时候，输出路程 if(temp.len==ans1) K++; else return temp.len; } if(ans[temp.v]>K) continue; for(int i=head[temp.v]; i!=-1; i=edge[i].next){ a_star n2; n2.v=edge[i].v; n2.len=edge[i].len+temp.len; q.push(n2); } } return -1; } int main(){ int i,j,k,l; int a,b,s; while(scanf("%d%d",&n,&m)!=EOF){ init(); while(m--){ scanf("%d%d%d",&a,&b,&s); insert(a,b,s); insert(b,a,s); } S=1,T=n,K=2; spfa(); ans1=dis[S]; // printf("%d\n",ans1); printf("%d\n",A_star()); } return 0; }
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原文地址：http://www.cnblogs.com/13224ACMer/p/4857566.html
