[Codeforces 545E] Paths and Trees

[题目链接]

        https://codeforces.com/contest/545/problem/E

[算法]

        首先求 u 到所有结点的最短路

        记录每个节点最短路径上的最后一条边

        答案即为以u为根的一棵最短路径生成树

        时间复杂度 : O(NlogN)

[代码]

       

#include
using namespace std;
const int MAXN = 3e5 + 10;
const long long INF = 1e60;

struct edge
{
        int to , w , nxt;
} e[MAXN << 1];

int n , m , s , tot;
int head[MAXN],u[MAXN],v[MAXN],w[MAXN],last[MAXN];
long long dist[MAXN];
bool visited[MAXN],vis[MAXN << 1];

template  inline void chkmax(T &x,T y) { x = max(x,y); }
template  inline void chkmin(T &x,T y) { x = min(x,y); }
template  inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v,int w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
}
inline void dijkstra(int s)
{
        priority_queue< pair<long long,int> , vector< pair<long long,int> > ,greater< pair<long long,int> > > q;
        for (int i = 1; i <= n; i++)
        {
                dist[i] = INF;
                visited[i] = false;        
        }        
        dist[s] = 0;
        q.push(make_pair(0,s));
        while (!q.empty())
        {
                int cur = q.top().second;
                q.pop();
                if (visited[cur]) continue;
                visited[cur] = true;
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to , w = e[i].w;
                        if (!visited[v] && dist[cur] + w <= dist[v])
                        {
                                if (dist[cur] + w < dist[v]) last[v] = i;
                                else if (w < e[last[v]].w) last[v] = i;
                                dist[v] = dist[cur] + w;
                                q.push(make_pair(dist[v],v));
                        }
                }
        }
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= m; i++)
        {
                read(u[i]); read(v[i]); read(w[i]);
                addedge(u[i],v[i],w[i]);
                addedge(v[i],u[i],w[i]); 
        } 
        read(s);
        dijkstra(s);
        for (int i = 1; i <= n; i++) vis[last[i]] = true;
        vector< int > ans;
        long long res = 0;
        for (int i = 2; i <= tot; i += 2)
        {
                if (vis[i] || vis[i - 1])
                {
                        ans.push_back(i >> 1);
                        res += e[i].w;
                }
        }
        printf("%I64d\n",res);
        for (unsigned i = 0; i < ans.size(); i++) 
                if (i == 0) printf("%d",ans[i]); 
                else printf(" %d",ans[i]);
        printf("\n");
        
        return 0;
    
}

 

转载于:https://www.cnblogs.com/evenbao/p/9745950.html