洛谷P2939 [USACO09FEB]改造路Revamping Trails

题意翻译

约翰一共有$N$)个牧场.由$M$条布满尘埃的小径连接.小径可以双向通行.每天早上约翰从牧场$1$出发到牧场$N$去给奶牛检查身体.

通过每条小径都需要消耗一定的时间.约翰打算升级其中$K$条小径，使之成为高速公路.在高速公路上的通行几乎是瞬间完成的，所以高速公路的通行时间为$0$.

请帮助约翰决定对哪些小径进行升级，使他每天从$1$号牧场到第$N$号牧场所花的时间最短

题目描述

Farmer John dutifully checks on the cows every day. He traverses some of the $M (1 <= M <= 50,000)$ trails conveniently numbered $1..M$ from pasture $1$ all the way out to pasture $N$ (a journey which is always possible for trail maps given in the test data). The $N$ ($1 <= N <= 10,000$) pastures conveniently numbered $1..N$ on Farmer John's farm are currently connected by bidirectional dirt trails. Each trail i connects pastures $P1_i$ and $P2_i$ ($1 <= P1_i <= N; 1 <= P2_i <= N$) and requires $T_i$ ($1 <= T_i <= 1,000,000$) units of time to traverse.

He wants to revamp some of the trails on his farm to save time on his long journey. Specifically, he will choose $K$ ($1 <= K <= 20$) trails to turn into highways, which will effectively reduce the trail's traversal time to $0$. Help FJ decide which trails to revamp to minimize the resulting time of getting from pasture $1$ to $N$.

TIME LIMIT: $2$ seconds

输入输出格式

输入格式：

Line $1$: Three space-separated integers: $N, M$, and $K$
Lines $2..M+1$: Line $i+1$ describes trail i with three space-separated integers: $P1_i, P2_i$, and $T_i$

输出格式：

Line $1$: The length of the shortest path after revamping no more than $K$ edges

输入输出样例

输入样例#1：

输出样例#1：

说明

$K$ is $1$; revamp trail $3->4$ to take time $0$ instead of $100$. The new shortest path is $1->3->4$, total traversal time now $1$.

思路：依旧是分层最短路，这次的分层最短是的$k$是有$k$次可以不花费任何代价去走一条路，那么我们把图分为$n$层，每条边连向映射点与原来边的起来的距离为$0$，然后跑$dijkstra$，就做完了。

代码：

#include<cstdio>

#include<cstring>

#include<queue>

#include<algorithm>

#include<cctype>

#define maxn 5000001

using namespace std;

int n,m,k,head[maxn],num,dis[maxn],s,t;

inline int qread() {

  char c=getchar();int num=0,f=1;

  for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;

  for(;isdigit(c);c=getchar()) num=num*10+c-'0';return num*f;

}

struct Edge {

  int v,w,nxt;

}e[maxn];

struct node {

  int x,y;

  bool operator < (const node &a) const {return y>a.y;}

};

inline void ct(int u, int v, int w) {

  e[++num].v=v;

  e[num].w=w;

  e[num].nxt=head[u];

  head[u]=num;

}

priority_queue<node>q;

inline void dijkstra() {

  memset(dis,0x3f,sizeof(dis));

  dis[1+n*k]=0;q.push((node){1+n*k,0});

  while(!q.empty()) {

  	int u=q.top().x,d=q.top().y;

  	q.pop();

  	if(d!=dis[u]) continue;

  	for(int i=head[u];i;i=e[i].nxt) {

  	  int v=e[i].v;

	  if(dis[v]>dis[u]+e[i].w) {

	  	dis[v]=dis[u]+e[i].w;

	  	q.push((node){v,dis[v]});

	  }

	}

  }

}

int main() {

  n=qread(),m=qread(),k=qread();

  for(int i=1,u,v,w;i<=m;++i) {

  	u=qread(),v=qread(),w=qread();

  	for(int j=0;j<=k;++j) {

  	  ct(u+j*n,v+j*n,w);

	  ct(v+j*n,u+j*n,w);

	  if(j) {

	  	ct(u+j*n,v+(j-1)*n,0);

	  	ct(v+j*n,u+(j-1)*n,0);

	  }

	}

  }

  dijkstra();

  int zrj=0x7fffffff;

  for(int i=0;i<=k;++i) zrj=min(zrj,dis[n+i*n]);

  printf("%d\n",zrj);

  return 0;

}