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
",ans);
}