Poj2449-BFS + astar

Find the nth short circuit. When I saw this question, I quickly thought of it, but I just couldn't implement it. After reading discuss, you can use Dijkstra to find the shortest path as the evaluation function and use astar to find the shortest path. The most common practice of astar is to use the priority queue to store nodes.

Use tpoint to save the nodes in astar, where to is the point, DIS represents the distance between the current node, and all is the total distance to the target.

#include<queue>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define maxn 1005#define inf 1<<30int sum1[maxn],sum2[maxn];int dis[maxn],cnt[maxn];bool vis[maxn];int tot1,tot2,n,m,t;int st,en,kth;struct TNode{int to,next,cost;bool operator<(const TNode &temp) const{return cost>temp.cost;}};struct TPoint{int to,dis,all;bool operator<(const TPoint &temp) const{return all>temp.all;}};vector<TNode>ans1,ans2;priority_queue<TNode>qu;priority_queue<TPoint>ans;void addedge(int u,int v,int w,bool flag){TNode tmp;tmp.to=u,tmp.cost=w;if(flag){tmp.next=sum1[v],sum1[v]=tot1++;ans1.push_back(tmp);}else{tmp.next=sum2[v],sum2[v]=tot2++;ans2.push_back(tmp);}}void dijkstra(){int k;fill(dis,dis+n+1,inf);memset(vis,0,sizeof(vis));while(!qu.empty()) qu.pop();TNode tmp1,tmp2;tmp1.to=en,tmp1.cost=0;qu.push(tmp1);dis[en]=0;while(!qu.empty()){tmp1=qu.top(),qu.pop();if(vis[tmp1.to]) continue;vis[tmp1.to]=1;for(k=sum1[tmp1.to];k!=-1;k=ans1[k].next){    if(dis[ans1[k].to]>dis[tmp1.to]+ans1[k].cost)    {        dis[ans1[k].to]=dis[tmp1.to]+ans1[k].cost;        tmp2.to=ans1[k].to;        tmp2.cost=dis[tmp2.to];        qu.push(tmp2);    }}}}int astar(){    int i,j,k;    TPoint tmp1,tmp2;    while(!ans.empty()) ans.pop();    memset(cnt,0,sizeof(cnt));    tmp1.to=st,tmp1.dis=0,tmp1.all=dis[st];    if(tmp1.all==inf) return -1;    ans.push(tmp1);    while(!ans.empty())    {        tmp1=ans.top(),ans.pop();        cnt[tmp1.to]++;        if(cnt[en]==kth) return tmp1.all;        if(cnt[tmp1.to]>kth) continue;        for(k=sum2[tmp1.to];k!=-1;k=ans2[k].next)        {            tmp2.to=ans2[k].to;            tmp2.dis=tmp1.dis+ans2[k].cost;            tmp2.all=tmp2.dis+dis[tmp2.to];            ans.push(tmp2);        }    }    return -1;}int main(){int i,u,v,w;while(scanf("%d%d",&n,&m)!=EOF){tot1=tot2=0;memset(sum1,-1,sizeof(sum1));memset(sum2,-1,sizeof(sum2));ans1.clear();ans2.clear();for(i=0;i<m;i++){scanf("%d%d%d",&u,&v,&w);addedge(u,v,w,1);addedge(v,u,w,0);}scanf("%d%d%d",&st,&en,&kth);dijkstra();if(st==en) kth++;printf("%d\n",astar());}return 0;}