BZOJ1598: [Usaco2008 Mar]牛跑步

传送门

K短路，普遍的算法是采用AStar求解，先建立反向边跑一遍dij,或者spfa什么的。跑出反向边的距离就可以看为估价函数中的$h()$。设$dist$为当前已经走过的距离，那么$f(node)=dist+h(son)$,然后跑一遍AStar，根据一些莫名其妙的东西可以得到第$i$次出栈的即为第$i$短路。

//BZOJ 1598
//by Cydiater
//2016.10.25
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <map>
#include <ctime>
#include <cmath>
#include <bitset>
#include <string>
#include <algorithm>
#include <iomanip>
using namespace std;
#define ll long long
#define up(i,j,n)		for(int i=j;i<=n;i++)
#define down(i,j,n)		for(int i=j;i>=n;i--)
#define pii pair<int,int>
#define mp make_pair
#define Pii priority_queue<pii,vector<pii>,greater<pii> >
const int MAXN=1e6+5;
const int oo=0x3f3f3f3f;
inline int read(){
	char ch=getchar();int x=0,f=1;
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int N,M,K,LINK[MAXN],undis[MAXN],len=0,f[MAXN],cnt[MAXN],ans[MAXN],dis[MAXN];
bool vis[MAXN];
struct edge{
	int y,next,v;
}e[MAXN];
Pii Q,q;
namespace solution{
	inline void insert(int x,int y,int v){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;e[len].v=v;}
	void init(){
		N=read();M=read();K=read();
		up(i,1,M){
			int x=read(),y=read(),v=read();
			insert(x,y,v);
			insert(y,x,-v);
		}
	}
	void pre_dijkstra(){
		memset(vis,0,sizeof(vis));
		memset(undis,10,sizeof(undis));
		undis[1]=0;Q.push(mp(undis[1],1));
		while(!Q.empty()){
			pii tmp=Q.top();Q.pop();
			int node=tmp.second;
			if(vis[node])continue;
			vis[node]=1;
			for(int i=LINK[node];i;i=e[i].next)
				if(undis[e[i].y]>undis[node]-e[i].v&&e[i].v<0){
					undis[e[i].y]=undis[node]-e[i].v;
					Q.push(mp(undis[e[i].y],e[i].y));
				}
		}
	}
	void now_AStar(){
		memset(cnt,0,sizeof(cnt));
		memset(ans,-1,sizeof(ans));
		q.push(mp(undis[N],N));
		while(!q.empty()){
			pii tmp=q.top();q.pop();
			int node=tmp.second;
			cnt[node]++;if(node==1)ans[cnt[node]]=tmp.first;
			for(int i=LINK[node];i;i=e[i].next)if(cnt[node]<=K&&e[i].v>0)
				q.push(mp(tmp.first-undis[node]+e[i].v+undis[e[i].y],e[i].y));
		}
	}
	void slove(){
		pre_dijkstra();
		now_AStar();
	}
	void output(){
		up(i,1,K)printf("%d\n",ans[i]);
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	init();
	slove();
	output();
	return 0;
}