ACM-ICPC 2018 南京赛区网络预赛 L Magical Girl Haze（Dijkstra变形）

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 and vv.

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

题意：给你一个带权有向图n(n<=1e5)个点m(m<=2e5)条边，你可以最多使k条边边权变为0，求1到n的最短路。（保证存在）

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思路：借用大佬思路和大佬代码。

最短路变形。用Dijkstra模板

dis[i][j]代表到i点，变j条边权值为0的最短路径

状态转移方程为

dis[v[i].to][j]=min(dis[v[i].to][j],dis[i][j]+v[i].distance);

dis[v[i].to][j+1]=min(dis[v[i].to][j+1],dis[i][j]);

代码：

#include<bits/stdc++.h>
#define maxn 200005
#define inf 0x3f3f3f3f3f3f3f3fLL
#define ll long long
using namespace std;
int n, m, k,head[200005],cnt;
struct node
{
    int to ,too;
    ll distance;
}v[200005];
void add(int uu,int vv,ll val)
{
    v[cnt].to=vv;
    v[cnt].too=head[uu];
    v[cnt].distance=val;
    head[uu]=cnt;
    cnt++;
}
bool book[100045][15];
ll dis[100405][15];
struct NNode
{
    int id,to,diss;
    bool operator<(const NNode &aa)const
    {
        return diss>aa.diss;
    }
};

void myinit()
{
    cnt=0;
    memset(head,-1,sizeof(head));
    memset(book, 0, sizeof(book));
    for(int i=0;i<=n;i++)
        for(int j=0;j<=k;j++)
        dis[i][j]=inf;
}
void SPFA(int s)
{
    priority_queue<NNode> q;
    NNode now,next;
    for(int i=0;i<=k;i++){
    now.to=s;
    now.id=i;
    now.diss=0;
    q.push(now);
    dis[s][i]=0;
    }
    int pos;
    while(!q.empty())
    {
        int point;
        now= q.top();
        q.pop();
        pos=now.id;
        point=now.to;
        if(book[point][now.id])
            continue;
        book[point][now.id]=1;
        for(int i = head[point]; i !=-1; i=v[i].too)
        {
            if(pos<k&&dis[point][pos] < dis[v[i].to][pos+1])
            {
                dis[v[i].to][pos+1] = dis[point][pos];
                if(!book[v[i].to][pos+1])
                {
                    next.id=pos+1;
                    next.to=v[i].to;
                    next.diss=dis[v[i].to][pos+1];
                    q.push(next);
                }
            }
            if(pos<=k&&v[i].distance + dis[point][pos] < dis[v[i].to][pos])
            {
                dis[v[i].to][pos] = dis[point][pos]  + v[i].distance;
                if(!book[v[i].to][pos])
                {
                    next.id=pos;
                    next.to=v[i].to;
                    next.diss=dis[v[i].to][pos];
                    q.push(next);
                }
            }
        }
    }
}
int x,y;
ll d;

int main()
{
    int i, j;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&k);
        myinit();
        for(i = 0; i < m; i++)
        {
            scanf("%d%d%lld", &x, &y, &d);
            add(x,y,d);
        }
        dis[1][0] = 0;
        SPFA(1);
        ll ans;
        ans=dis[n][0];
        for( i=0; i<=k; i++)
        {
            ans=min(ans,dis[n][i]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}