free【分层图最短路】

free

传送门   来源: 牛客网

题目描述

Your are given an undirect connected graph.Every edge has a cost to pass.You should choose a path from S to T and you need to pay for all the edges in your path. However, you can choose at most k edges in the graph and change their costs to zero in the beginning. Please answer the minimal total cost you need to pay.

输入描述:

The first line contains five integers n,m,S,T,K.

For each of the following m lines, there are three integers a,b,l, meaning there is an edge that costs l between a and b.

n is the number of nodes and m is the number of edges.

输出描述:

An integer meaning the minimal total cost.

输入

3 2 1 3 1
1 2 1
2 3 2

输出

1

备注:

1≤n,m≤103,1≤S,T,a,b≤n,0≤k≤m,1≤l≤106.

Multiple edges and self loops are allowed.

题目描述：

给出n个点，和m条带权边，并且可以选定几条边令其权值为0，但选择的边数最多是k条，求s和t两点的最短距离。

思路：

有k次机会使边的权值为0，是分层最短路的经典问题。

具体方法看下图：(图中序号是从0开始的，下面讲解用从1开始的)

图源：sugarbliss

根据上图可以知道当k=2时，需要建k+1层的图，其中第一层序号是[1,n]，往下依次是[1+n,n+n]、[1+2n+n+2*n]

在使用链式前向星存图的时候，要同时存下一列的权值，并将上下两层的权值设为0。

最终最短路的长度出现在每一行的最后，即：n、2*n、3*n。

例如（1，5）=10 要存（1+5，5+5）=10、（1+2*5，5+2*5）=1

（1，5+5+1）=0   等。

（看不懂直接看代码很好理解，找了一下午才找了个感觉好适应的板子）

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAX=5e6;
const int INF=0x3f3f3f3f;
LL n,m,s,t,k;
LL head[MAX+5],ans;
LL dis[MAX+5],vis[MAX+5];
struct note{
    LL to;
    LL len;
    LL next;
}edge[MAX+5];
void addedge(LL u,LL v,LL w)
{
    edge[ans].to=v;
    edge[ans].len=w;
    edge[ans].next=head[u];
    head[u]=ans++;
}
void init()
{
    memset(head,-1,sizeof(head));
    ans=0;
}
struct node{
    LL u,len;
    node(LL a,LL b){
        u=b;
        len=a;
    }
    friend bool operator < (node a,node b){
        return a.len>b.len;
    }
};
priority_queue<node>q;
void diji(LL s)
{
    for(LL i=0;i<=(k+1)*n;i++){
        dis[i]=INF;
        vis[i]=0;
    }
    dis[s]=0;
    q.push(node(0,s));
    while(!q.empty()){
        int k=q.top().u;
        q.pop();
        if(vis[k]){
            continue;
        }
        vis[k]=1;
        for(LL i=head[k];~i;i=edge[i].next){
            int t=edge[i].to;
            if((dis[t]>dis[k]+edge[i].len&&edge[i].len!=INF+2)){
                dis[t]=dis[k]+edge[i].len;
                q.push(node(dis[t],t));
            }
        }
    }
}
int main()
{
    scanf("%lld%lld%lld%lld%lld", &n, &m, &s,&t,&k);
    init();
    while(m--)
    {
        LL u, v, w;
        scanf("%lld%lld%lld",&u, &v, &w);
        for(LL i = 0; i <= k; i++)
        {
            addedge(u + i * n, v + i * n, w);
            addedge(v + i * n, u + i * n, w);
            if(i != k)
            {
                addedge(u + i * n, v + (i + 1) * n, 0);
                addedge(v + i * n, u + (i + 1) * n, 0);
            }
        }
    }
    diji(s);
    LL ans = INF;
    for(LL i = 0; i <= k; i++)
        ans = min(ans,dis[t + i * n]);
    printf("%lld\n",ans);
}