【 Educational Codeforces Round 54 (Rated for Div. 2) D. Edge Deletion】dij+思维

D. Edge Deletion

题意

$在一个n个点m条边的无向图中起点为1，设初始到达第i个点的最短距离为d[i]$
$现在要求在图上删边，使剩下的边不超过k条，并让尽量多的点d[i]与之前相等$

做法

$我们发现dij跑出来的图最终每个点只有一个前驱$
$那么这就是一棵树，在树上从根节点往下保存边不会影响子孙的d$
$所以跑出dij树之后从根往下保留k条边就可以了。$

坑点

$最短路用dij不要用spfa$
$边权和会超过longlong$

代码

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include 
#include
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;

const int maxn = 3e5+5;
const int Mod=1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const double e=exp(1);
const db PI = acos(-1);
const db ERR = 1e-10;

#define Se second
#define Fi first
#define pb push_back
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<#define dbg3(x1,x2,x3) cout<<#x1<<" = "<typedef long long ll;
#define maxm 600005
#define maxn 300005
#define inf 0x3f3f3f3f3f3f3f3f
struct P
{
    int to;
    ll cost;
    bool operator < (const P & a) const
    {
        return cost>a.cost;
    }
};
struct node
{
    int to;
    ll val;
    int nxt;
    int id;
    int s;
}edge[maxm];
int head[maxn],tot;//head和tot记得重置，head重置为-1
int n,m;//点数，边数，不要再main里面重新定义
bool vis[maxn];//每次在dij里面初始化为0
ll dis[maxn];//根据题意初始化为inf可能int可能longlong
void addedge(int x,int y,ll val,int id)
{
    edge[tot].to=y;
    edge[tot].val=val;
    edge[tot].nxt=head[x];
    edge[tot].id=id;
    head[x]=tot++;
}
int pre[maxn],prem[maxn];
int vv[maxn];
void Dijkstra(int s)
{
    memset(vis,0,sizeof(vis));
    fill(dis,dis+n+2,inf);
    dis[s]=0;
    priority_queue<P>q;
    q.push(P{s,0});
    while(!q.empty())
    {
        P p1=q.top();q.pop();
        int u=p1.to;
        if(vis[u])continue;
        vis[u]=1;
        for(int i=head[u];i+1;i=edge[i].nxt)
        {
            int v=edge[i].to;
            if(vis[v])continue;
            if(dis[v]>dis[u]+edge[i].val)
            {
                pre[v]=u;//记录每个点的点前驱
                prem[v]=edge[i].id;//记录每个点的边前驱
                dis[v]=dis[u]+edge[i].val;
                q.push(P{v,dis[v]});
            }
        }
    }
}
vector<pii> G[maxn];
vector<int> ans;
int ind[maxn];
int cc,k;
void dfs(int st)
{
    if(cc==k) return ;
    for(int i=0;i<G[st].size();i++)
    {
        cc++;
        ans.push_back(G[st][i].Se);
        if(cc==k) return;
        dfs(G[st][i].Fi);
        if(cc==k) return ;
    }
}
int main()
{
   int u,v;
   ll w;
   memset(head,-1,sizeof(head));
   scanf("%d%d%d",&n,&m,&k);
   for(int i=1;i<=m;i++)
   {
        scanf("%d%d%lld",&u,&v,&w);
        addedge(u,v,w,i);
        addedge(v,u,w,i);
   }
   Dijkstra(1);
   if(k>=n-1)
   {
       printf("%d\n",n-1);
       for(int i=2;i<=n;i++)
       {
           printf("%d ",prem[i]);
       }
   }
   else
   {
       int re=n-1;
       for(int i=2;i<=n;i++)
       {
          G[pre[i]].push_back(pii(i,prem[i]));//重新根据前驱生成一棵树
       }
        cc=0;
        dfs(1);
        printf("%d\n",k);
        for(int i=0;i<ans.size();i++)
        {
            printf("%d ",ans[i]);
        }
   }
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

---