D. Edge Deletion
time limit per test
2.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an undirected connected weighted graph consisting of nn vertices and mm edges. Let's denote the length of the shortest path from vertex 11 to vertex ii as didi .
You have to erase some edges of the graph so that at most kk edges remain. Let's call a vertex ii good if there still exists a path from 11 to ii with length didi after erasing the edges.
Your goal is to erase the edges in such a way that the number of good vertices is maximized.
Input
The first line contains three integers nn , mm and kk (2≤n≤3⋅1052≤n≤3⋅105 , 1≤m≤3⋅1051≤m≤3⋅105 , n−1≤mn−1≤m , 0≤k≤m0≤k≤m ) — the number of vertices and edges in the graph, and the maximum number of edges that can be retained in the graph, respectively.
Then mm lines follow, each containing three integers xx , yy , ww (1≤x,y≤n1≤x,y≤n , x≠yx≠y , 1≤w≤1091≤w≤109 ), denoting an edge connecting vertices xx and yy and having weight ww .
The given graph is connected (any vertex can be reached from any other vertex) and simple (there are no self-loops, and for each unordered pair of vertices there exists at most one edge connecting these vertices).
Output
In the first line print ee — the number of edges that should remain in the graph (0≤e≤k0≤e≤k ).
In the second line print ee distinct integers from 11 to mm — the indices of edges that should remain in the graph. Edges are numbered in the same order they are given in the input. The number of good vertices should be as large as possible.
Examples
Input
Copy
3 3 2 1 2 1 3 2 1 1 3 3
Output
Copy
2 1 2
Input
Copy
4 5 2 4 1 8 2 4 1 2 1 3 3 4 9 3 1 5
Output
Copy
2 3 2
題意：給你一個無向圖，求出1到其他點的最短路di，讓你保留最多k條邊其他的邊刪除，使得最多得點di不發生變化
思路：先求出1點到其他點的最短路，標記最短路的路徑，然後從1點開始dfs或bfs找出k條或者n-1邊就return；
反思：思路很好想，很容易想到先找對短路在找min(n-1,k)條邊就好，但是無辜的我在64組測試數據T了無數次，最後毫無辦法莽了一發dij標記在隊（其實是dij不需要標記在隊的，只是spfa才需要的，此題spfa被卡點了），然後跑的賊幾把快300ms這題還給的2500ms，無語。
dij標記在隊代碼：
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#define ll long long
#define qq printf("QAQ\n");
using namespace std;
const int maxn=3e5+5;
const int inf=0x3f3f3f3f;
const ll linf=0x3f3f3f3f3f3f3f3f;
const int mod=1e9+7;
const double e=exp(1.0);
const double pi=acos(-1);
const double eps=1e-6;
struct Edge{
int id,to,w,next;
}edge[maxn<<1];
struct node{
int pos;
ll w;
bool operator < (const node &t)const{
return w > t.w;
}
};
int n,m,k,cnt,head[maxn],last[maxn],ans[maxn];
ll dis[maxn];
bool use[maxn];
void addedge(int st,int en,int id,int w)
{
edge[cnt].to=en;
edge[cnt].w=w;
edge[cnt].id=id;
edge[cnt].next=head[st];
head[st]=cnt++;
}
void dij()
{
for(int i=1;i<=n;i++)dis[i]=linf,last[i]=-1;
dis[1]=0ll;
priority_queue<node>q;
q.push((node){1,0ll});
use[1]=1;
while(!q.empty())
{
node now=q.top();
q.pop();
use[now.pos]=0;
for(int i=head[now.pos];i!=-1;i=edge[i].next)
{
int next=edge[i].to;
if(dis[next]>dis[now.pos]+(ll)edge[i].w)
{
dis[next]=dis[now.pos]+(ll)edge[i].w;
if(!use[next])q.push((node){next,dis[next]}),use[next]=1;
last[next]=now.pos;
}
}
}
}
int num;
void bfs()
{
queue<int>q;
q.push(1);
if(num==n||num==k)return ;
while(!q.empty())
{
int now=q.front();
q.pop();
for(int i=head[now];i!=-1;i=edge[i].next)
{
if(last[edge[i].to]==now){
ans[num++]=edge[i].id;
q.push(edge[i].to);
if(num==n||num==k)return ;
}
}
}
}
int main()
{
scanf("%d%d%d",&n,&m,&k);
int st,en,w;
//memset(head,-1,sizeof head);
for(int i=1;i<=n;i++)head[i]=-1,use[i]=0;
cnt=0;
for(int i=1;i<=m;i++)
{
scanf("%d%d%d",&st,&en,&w);
addedge(st,en,i,w);
addedge(en,st,i,w);
}
num=0;
dij();
bfs();
printf("%d\n",num);
for(int i=0;i<num;i++)
printf("%d%c",ans[i],i==num-1?'\n':' ');
return 0;
}