poj2449---BFS+Astar

寻找第k短路。看到这道题时，很快就想到了思路，但就是实现不了。看了discuss，可以先用dijkstra求出最短路作为估价函数，在astar来实现求出第k短路，而astar最惯常做法是用优先队列保存节点。

用TPoint来保存astar里的节点，其中to是指向的位置，dis代表当前节点中已经过的距离，all是到目标的总距离。

#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;}