Magical Girl Haze（迪杰斯特拉+dp）

本文链接：https://blog.youkuaiyun.com/xigongdali/article/details/82524663

There are NN cities in the country, and MMdirectional roads from uu to v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become 00. Now she wants to go to City NN, please help her calculate the minimum distance.

Input

The first line has one integer T(1 \le T\le 5)T(1≤T≤5), then following TT cases.

For each test case, the first line has three integers N, MN,M and KK.

Then the following MM lines each line has three integers, describe a road, U_i, V_i, C_iUi,Vi,Ci. There might be multiple edges between uu andvv.

It is guaranteed that N \le 100000, M \le 200000, K \le 10N≤100000,M≤200000,K≤10,
0 \le C_i \le 1e90≤Ci≤1e9. There is at least one path between City 11 and City NN.

Output

For each test case, print the minimum distance.

样例输入复制

样例输出复制

题目来源

ACM-ICPC 2018 南京赛区网络预赛

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <queue>
#define INF 0x3f3f3f3f3f3f3f3f
#define MAXN 200000
using namespace std;
/*  Dijkstar 算法+堆优化 使用优先队列优化，复杂度O(ElogE)
*   使用优先队列优化Dijkstra算法
*	复杂度O(ElogE)
*	注意对vector<Edge>E[MAXN]进行初始化后加边
*/
long long  dis[MAXN][20];
bool vis[MAXN][20];
struct qnode          //这是为优先队列服务的结构体
{
    int u;
    long long c;      //c表示的dis[i]，即起点到i的距离
    int num;
    qnode(int _u=0,long long _c=0,int _num=0):u(_u),c(_c),num(_num){}
    bool operator <(const qnode &r)const  //c起排序作用，最短的先。
    {
        if(c!=r.c)  return c>r.c;
        else        return num>r.num;
    }
};
struct Node
{
   int v;
   long long w;  //w是指两点的距离
   Node(int _v=0,long long _w=0):v(_v),w(_w){}
};
vector<Node> G[MAXN];
void init(int n)
{
     int i;
     for(i=0;i<=n;i++)
     G[i].clear();
     memset(vis,false,sizeof(vis));
     memset(dis,INF,sizeof(dis));
}
void addedge(int u,int v,int w)
{
    G[u].push_back(Node(v,w));
}
void dijkstra(int start,int n,int k) //点的编号从1开始
{
     int i;
     priority_queue<qnode>  que;
     while(!que.empty())
     que.pop();                //对于多组测试数据的话，就要不断地清空
     dis[start][0]=0;
     que.push(qnode(start,0,0));
     qnode tmp;
     while(!que.empty())
     {
          tmp=que.top();
          que.pop();
          int u=tmp.u;
          int num=tmp.num;
          if(vis[u][num])    continue;  //因为是双向图嘛，所以vis标记是关键的
          vis[u][num]=true;             //这个vis标记的是当前在哪个结点，算的是起点到当前这个点的最短距离。
                                   //当以这点为转达时就标记。因为他的最短距离已经在之前算过了，之前一定是最短
          for(i=0;i<G[u].size();i++)
          {
              int         v=G[u][i].v;
              long long   w=G[u][i].w;
              if(!vis[v][num]&&dis[v][num]>dis[u][num]+w)    //考虑同num状态    ，因为总共就两种状态转移，对当前边取与不取0的讨论
              {                                              //不取意味着num不变，取意味着num+1
                   dis[v][num]=dis[u][num]+w;
                   que.push(qnode(v,dis[v][num],num));
              }
              if(!vis[v][num+1]&&(num<k)&&(dis[v][num+1]>dis[u][num]))   //考虑同num+1状态(转态转移时考虑的是 dis[u][num-1]与dis[v][num]的比较
              {                                                          //但算法要写num+1的
                  dis[v][num+1]=dis[u][num];
                  que.push(qnode(v,dis[v][num+1],num+1));
              }
          }
     }
}
int main()
{
    int n,m,k;
    int i;
    int u,v;
    long long w;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        init(n);
        for(i=0;i<m;i++)
        {
            scanf("%d%d%lld",&u,&v,&w);
            addedge(u,v,w);
        }
        dijkstra(1,n,k);
        printf("%lld\n",dis[n][k]);
    }
    return 0;
}