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+5,maxm=2e5+5;
ll d[11][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({0,1,0});
    d[0][1]=0;
    while(!q.empty())
    {
        auto u=q.top();q.pop();
        if(u.v==n)return;
        for(int i=head[u.v];i!=-1;i=edges[i].next)
        {
            edge &v=edges[i];
            if(u.k<k&&d[u.k+1][v.v]>d[u.k][u.v])
                q.push({d[u.k+1][v.v]=d[u.k][u.v],v.v,u.k+1});
            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,-1,sizeof head);
        cnt=0;
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0,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=0;i<=k;i++)
            minn=min(minn,d[i][n]);
        printf("%lld
",minn);
    }
    return 0;
}