ACM-ICPC 2018 南京赛区网络预赛 L.Magical Girl Haze(分层最短路)

There are N cities in the country, and M directional roads from u to v(1≤u,v≤n). Every road has a distance c_i. Haze is a Magical Girl that lives in City 1, she can choose no more than K roads and make their distances become 0. Now she wants to go to City N, please help her calculate the minimum distance.

Input
The first line has one integer T(1≤T≤5), then following T cases.
For each test case, the first line has three integers N,M and K.
Then the following M lines each line has three integers, describe a road, U_i,V_i,C_i​. There might be multiple edges between u and v.
It is guaranteed that N≤100000,M≤200000,K≤10,0≤C_i≤1e9. There is at least one path between City 1 and City N.

Output
For each test case, print the minimum distance.

样例输入
1
5 6 1
1 2 2
1 3 4
2 4 3
3 4 1
3 5 6
4 5 2

样例输出
3

题意

给你一个图，问你从1到N最多让K条边为0的最短路

题解

d[i][j]代表到j点已经删去i条边的最小值

跑一遍dij

代码

 #include<bits/stdc++.h>
 using namespace std;

 #define ll long long
 const int maxn=1e5+,maxm=2e5+;
 ll d[][maxn];
 int n,m,k;
 int head[maxn],cnt;
 struct edge
 {
     int v,next;
     ll w;
 }edges[maxm];
 struct node
 {
     ll w;
     int v,k;
     bool operator<(const node &D)const{
         return w>D.w;
     }
 };
 void dij()
 {
     memset(d,0x3f,sizeof d);
     priority_queue<node>q;
     q.push({,,});
     d[][]=;
     while(!q.empty())
     {
         auto u=q.top();q.pop();
         if(u.v==n)return;
         for(int i=head[u.v];i!=-;i=edges[i].next)
         {
             edge &v=edges[i];
             if(u.k<k&&d[u.k+][v.v]>d[u.k][u.v])
                 q.push({d[u.k+][v.v]=d[u.k][u.v],v.v,u.k+});
             if(d[u.k][v.v]>d[u.k][u.v]+v.w)
                 q.push({d[u.k][v.v]=d[u.k][u.v]+v.w,v.v,u.k});
         }
     }
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         memset(head,-,sizeof head);
         cnt=;
         scanf("%d%d%d",&n,&m,&k);
         for(int i=,u,v,w;i<m;i++)
         {
             scanf("%d%d%d",&u,&v,&w);
             edges[cnt].v=v;
             edges[cnt].w=1LL*w;
             edges[cnt].next=head[u];
             head[u]=cnt++;
         }
         dij();
         ll minn=0x3f3f3f3f3f3f3f3f;
         for(int i=;i<=k;i++)
             minn=min(minn,d[i][n]);
         printf("%lld\n",minn);
     }
     return ;
 }