bzoj 1598（SPFA+A*）

传送门

~~~A*的经典应用~~~

题解：

先预处理反向图的单源(T)最短路，之后启发式搜索，每次从优先队列中弹出f(p)最小的点p,如果点p就是t,则计算t出队的次数，如果当前为t的第k次出队,则当前路径的长度就是s到t的第k短路的长度，算法结束，否则遍历与p相连的所有的边,将扩展出的到p的邻接点信息加入到优先队列继续搜。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<vector>using namespace std;typedef long long ll;const int MAXN=1004;int n,m,K;ll dis[MAXN],ans[MAXN];int vis[MAXN];vector<int> g[MAXN],rg[MAXN];vector<int> l[MAXN],rl[MAXN];struct A {int id;ll g,h;A(int _id=0,ll _g=0,ll _h=0):id(_id),g(_g),h(_h){}friend bool operator <(const A &a,const A &b) {return a.g+a.h>b.g+b.h;}};inline int read() {int x=0;char c=getchar();while (c<'0'||c>'9') c=getchar();while (c>='0'&&c<='9') x=x*10+c-'0',c=getchar();return x;}inline void SPFA(int source) {queue<int > q;memset(dis,0x3f,sizeof(dis));memset(vis,0,sizeof(vis));q.push(source),dis[source]=0,vis[source]=true;while (!q.empty()) {int p=q.front();q.pop();vis[p]=0;for (int i=0;i<rg[p].size();++i) {int v=rg[p][i];if (dis[v]>dis[p]+rl[p][i]) {dis[v]=dis[p]+rl[p][i];if (!vis[v]) vis[v]=1,q.push(v);}}}}inline void Astar(int source,int sink) {priority_queue<A > q;q.push(A(source,0,dis[source]));int tim=0;while (!q.empty()) {A cur=q.top();q.pop();if (++vis[cur.id]>K) continue;if (cur.id==sink) ans[++tim]=cur.g;if (vis[sink]==K) return ;for (int i=0;i<g[cur.id].size();++i) {int v=g[cur.id][i];q.push(A(v,cur.g+l[cur.id][i],dis[v]));}}}int main() {//freopen("bzoj 1598.in","r",stdin);n=read(),m=read(),K=read();for (int i=1;i<=n;++i) g[i].clear(),rg[i].clear(),l[i].clear(),rl[i].clear();for (register int i=0;i<m;++i) {int u=read(),v=read(),w=read();g[u].push_back(v),rg[v].push_back(u);l[u].push_back(w),rl[v].push_back(w);}SPFA(1);memset(ans,-1,sizeof(ans));Astar(n,1);for (int i=1;i<=K;++i) printf("%lld\n",ans[i]);return 0;}