Educational Codeforces Round 54 (Rated for Div. 2) 1076D. Edge Deletion

题目链接：http://codeforces.com/problemset/problem/1076/D

题意：一个  n 个顶点 m 条边的无向图，定义顶点 v 到 顶点 1 的最短距离 di，现在最多留下 k 条边，如果一个顶点，在新遗留的图中到顶点 1 的最短距离还是 di，则称这个点为好顶点，问在保证好顶点最多的情况下，应该讲哪些边留下来。 输出留下的边数以及输入时这些边的编号。

思路：最短路树：在跑最短路的过程中，出掉多余的边，留下来的是一颗 n - 1 条边的树，称之为最短路树。而在最短路树上的节点一定是好顶点，所以遗留下来的边数为 e = min(k，n-1)。dijstra 算法在松弛的过程中，用数组 f[v] = u，将这条边以及它的序号保留下来。然后再将这些边重新建图-成一颗树，广度遍历取 e 条边。 

AC代码：

#include<bits/stdc++.h>

using namespace std;
#define per(i,a,n) for (int i=a;i<n;i++)
#define rep(i,a,n) for (int i=n-1;i>=a;i--)
#define runfile freopen("E:/AllCode/codeblocks project/Root/data.txt", "r", stdin)
#define stopfile fclose(stdin)
#define tt long long
#define MAX 100050
typedef long long ll;
const int maxn = 300005;
const int maxm = 300005;
const long long INF = LONG_LONG_MAX;
const int MOD = 998244353;

struct edge
{
    int u,v,order,next;
    long long w;
    edge(){}
    edge(int _u, int _v, long long _w, int _order, int _next) : u(_u), v(_v), w(_w), order(_order), next(_next){}
}g[2*maxm],tree[2*maxm];

struct node
{
    int v;
    long long w;
    bool operator < (const node &a) const
    {
        return w > a.w;
    }
}dis[maxn];

struct father
{
    int u;
    int order;
}f[maxn];

int n,m,k,s,e,cnt,cnt1,first[maxn],first1[maxn],vis[maxn],ans[maxm];

void init()
{
    memset(vis, 0, sizeof(vis));//mark the edges
    for(int i = 1; i <= n; i++)
    {
        dis[i].v = i;
        dis[i].w = INF;
        first[i] = -1;
        first1[i] = -1;
        f[i].u = i;
    }
    cnt = 1;
    cnt1 = 1;
    s = 1;
    e = min(n-1, k);
}

void add_edge(int u, int v, long long w)
{
    g[cnt] = edge(u, v, w, cnt, first[u]);
    first[u] = cnt;

    //undirected graph
    int temp = cnt + m;
    g[temp] = edge(v, u, w, temp, first[v]);
    first[v] = temp;
    cnt++;
}

void add_edge1(int u, int v, long long w, int o)
{
    tree[cnt1] = edge(u, v, w, o, first1[u]);
    first1[u] = cnt1;
    cnt1++;
}

void dijstra_tree(int s)
{
    priority_queue<node> pq;
    dis[s].v = s;
    dis[s].w = 0;
    pq.push(dis[s]);
    while(!pq.empty())
    {
        node now = pq.top();
        pq.pop();
        int u = now.v;
        if(vis[u])  continue;
        vis[u] = true;
        for(int i = first[u]; i != -1; i = g[i].next)
        {
            int v = g[i].v;
            long long w = g[i].w;
            int o = g[i].order > m ? g[i].order - m : g[i].order;
            if(!vis[v] && (w + dis[u].w < dis[v].w))
            {
                dis[v].w = w + dis[u].w;
                f[v].u = u;//save father node
                f[v].order = o;//save start order
//                add_edge1(u, v, w, o);
                pq.push(dis[v]);
            }
        }
    }
    //create tree
    for(int i = 1; i <= n; i++)
    {
        if(f[i].u != i)
        {
            add_edge1(f[i].u, i, 0, f[i].order);
        }
    }
}

int delete_edge()
{
    queue<int> q;
    q.push(1);
    int j = 0;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = first1[u]; i != -1; i = tree[i].next)
        {
            if(j >= e)   break;
            ans[j] = tree[i].order;
            j++;
            q.push(tree[i].v);
        }
    }
    return j;
}

int main()
{
    //runfile;
    ios::sync_with_stdio(false);
    int from, to, weight;
    while(cin>>n>>m>>k)
    {
        init();
        for(int i = 0; i < m; i++)
        {
            cin>>from>>to>>weight;
            add_edge(from, to, weight);
        }

        dijstra_tree(s);
        int j = delete_edge();

        printf("%d\n",e);
        if(e != 0)
        {
            for(int i = 0; i < j-1; i++)
                printf("%d ",ans[i]);
            printf("%d\n",ans[j-1]);
        }
    }
    //stopfile;
    return 0;
}