[USACO09FEB]改造路Revamping Trails

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

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，使他每天早上到牧场W花的时间最少.输出这个最少 的时间.

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：

4 4 1 
1 2 10 
2 4 10 
1 3 1 
3 4 100 

输出样例#1：

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.

建立分层图。
f[u][t]表示在节点u时已经免费乘坐t次的最少花
费。照样跑最短路。
枚举与u相连的所有节点v,w(u,v)表示权值。
若t<k:
f[v][t+1]=min(f[v][t+1],f[u][t])
对于所有:
f[v][t]=min(f[v][t],f[u][t]+w(u,v))

不过这题数据大一些，要用优先队列优化SPFA

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct Node
{
  int next,to,dis;
}edge[100001];
struct XXX
{
  int dis,x,k;
  friend bool operator<(XXX a,XXX b)
  {return a.dis>b.dis;}
};
int num,head[100001];
int dist[100001][21];
int n,m,k,S,T;
int ans;
bool vis[100001][21];
void add(int u,int v,int d)
{
  num++;
  edge[num].next=head[u];
  head[u]=num;
  edge[num].to=v;
  edge[num].dis=d;
}
void SPFA()
{int i;
  priority_queue<XXX>Q;
  Q.push((XXX){0,S,0});
  dist[S][0]=0;
  while (Q.empty()==0)
  {
    XXX u=Q.top();
    Q.pop();
    for (i=head[u.x];i;i=edge[i].next)
      {int v=edge[i].to;
    if (dist[v][u.k]>dist[u.x][u.k]+edge[i].dis)
      {
        dist[v][u.k]=dist[u.x][u.k]+edge[i].dis;
        Q.push((XXX){dist[v][u.k],v,u.k});
      }
    if (u.k+1<=k&&dist[v][u.k+1]>dist[u.x][u.k])
      {
        dist[v][u.k+1]=dist[u.x][u.k];
        Q.push((XXX){dist[v][u.k+1],v,u.k+1});
      }
      }
  }
}
int main()
{int i,u,v,c;
  cin>>n>>m>>k;
  memset(dist,127/2,sizeof(dist));
  S=1;T=n;
  for (i=1;i<=m;i++)
    {
      scanf("%d%d%d",&u,&v,&c);
      add(u,v,c);
      add(v,u,c);
    }
  SPFA();
  cout<<dist[n][k];
}