HDU4871-Shortest-path tree

Shortest-path tree

                                                                    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 130712/130712 K (Java/Others)
                                                                                            Total Submission(s): 1674    Accepted Submission(s): 566

Problem Description

Given a connected, undirected graph G, a shortest-path tree rooted at vertex v is a spanning tree T of G, such that the path distance from root v to any other vertex u in T is the shortest path distance from v to u in G. 
We may construct a shortest-path tree using the following method:
We consider a shortest-path tree rooted at node 1. For every node i in the graph G, we choose a shortest path from root to i. If there are many shortest paths from root to i, we choose the one that the sequence of passing nodes' number is lexicographically minimum. All edges on the paths that we chose form a shortest-path tree.
Now we want to know how long are the longest simple paths which contain K nodes in the shortest-path tree and how many these paths? Two simple paths are different if the sets of nodes they go through are different.

 

Input

The first line has a number T (T <= 10), indicating the number of test cases.
For each test case, the first line contains three integers n, m, k(1<=n<=30000,1<=m<=60000,2<=k<=n), denote the number of nodes, the number of edges and the nodes of required paths.
Then next m lines, each lines contains three integers a, b, c(1<=a, b<=n, 1<=c<=10000),denote there is an edge between a, b and length is c.

 

Output

For each case, output two numbers, denote the length of required paths and the numbers of required paths.

 

Sample Input

16 6 41 2 12 3 13 4 12 5 13 6 15 6 1

 

Sample Output

3 4

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

这题和HYSBZ4016最短路径树问题一模一样，具体可以看我的HYSBZ4016的题解

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <cmath>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;const int maxn = 2e5 + 10;int n, m, K, cnt1, cnt2;char ch;int s1[maxn], nt1[maxn], e1[maxn], v1[maxn];int s2[maxn], nt2[maxn], e2[maxn], v2[maxn];int vis[maxn], dis[maxn];int cnt[maxn], mx[maxn], ans1, ans2;int dfs(int k, int fa, int sum){int ans = mx[k] = 0;cnt[k] = 1;for (int i = s2[k]; i != -1; i = nt2[i]){if (vis[e2[i]] || e2[i] == fa) continue;int y = dfs(e2[i], k, sum);if (mx[y] < mx[ans]) ans = y;cnt[k] += cnt[e2[i]];mx[k] = max(mx[k], cnt[e2[i]]);}mx[k] = max(mx[k], sum - cnt[k]);return mx[k] < mx[ans] ? k : ans;}int deep(int k, int fa, int dep){if (dep == K) return K - 1;int ans = dep;mx[dep] = cnt[dep] = 0;for (int i = s2[k]; ~i; i = nt2[i]){if (vis[e2[i]] || e2[i] == fa) continue;ans = max(ans, deep(e2[i], k, dep + 1));}return ans;}void get(int k, int fa, int dep, int len, int sum){if (dep == K) return;if (K - 1 - dep <= sum && cnt[K - 1 - dep]){if (ans1 < len + mx[K - 1 - dep]){ans1 = len + mx[K - 1 - dep];ans2 = cnt[K - 1 - dep];}else if (ans1 == len + mx[K - 1 - dep]) ans2 += cnt[K - 1 - dep];}for (int i = s2[k]; ~i; i = nt2[i]){if (vis[e2[i]] || e2[i] == fa) continue;get(e2[i], k, dep + 1, len + v2[i], sum);}}void put(int k, int fa, int dep, int len, int sum){if (dep == K) return;if (mx[dep] < len) mx[dep] = len, cnt[dep] = 1;else if (mx[dep] == len) cnt[dep]++;for (int i = s2[k]; ~i; i = nt2[i]){if (vis[e2[i]] || e2[i] == fa) continue;put(e2[i], k, dep + 1, len + v2[i], sum);}}void Find(int k){int len = deep(k, k, 0);if (len + len + 1 < K) return;cnt[0] = 1;for (int i = s2[k]; ~i; i = nt2[i]){if (vis[e2[i]]) continue;get(e2[i], k, 1, v2[i], len);put(e2[i], k, 1, v2[i], len);}}void work(int k, int sum){mx[0] = INF;int y = dfs(k, k, sum);vis[y] = 1, Find(y);for (int i = s2[y]; ~i; i = nt2[i]){if (vis[e2[i]]) continue;if (cnt[e2[i]] < cnt[y]) work(e2[i], cnt[e2[i]]);else work(e2[i], sum - cnt[y]);}}struct Edge{int u, v, w;bool operator<(const Edge &a)const{return u == a.u ? v > a.v:u < a.u;}}p[maxn];struct node{int id, dis;bool operator<(const node& a)const{return dis == a.dis ? id > a.id : dis > a.dis;}}pre, nt;void dfs(int k){vis[k] = 0;for (int i = s1[k]; ~i; i = nt1[i]){if (dis[e1[i]] == dis[k] + v1[i]){if (!vis[e1[i]]) continue;nt2[cnt2] = s2[k], s2[k] = cnt2, e2[cnt2] = e1[i], v2[cnt2++] = v1[i];nt2[cnt2] = s2[e1[i]], s2[e1[i]] = cnt2, e2[cnt2] = k, v2[cnt2++] = v1[i];dfs(e1[i]);}}}void Dijkstra(){priority_queue<node> q;memset(dis, INF, sizeof dis);memset(vis, 0, sizeof vis);pre = { 1,dis[1] = 0 };q.push(pre);while (!q.empty()){pre = q.top();q.pop();vis[pre.id] = 1;for (int i = s1[pre.id]; ~i; i = nt1[i]){if (vis[e1[i]]) continue;if (dis[e1[i]] > pre.dis + v1[i]){nt = { e1[i],dis[e1[i]] = pre.dis + v1[i] };q.push(nt);}}}}int main(){int t;scanf("%d", &t);while (t--){scanf("%d%d%d", &n, &m, &K);cnt1 = cnt2 = ans1 = ans2 = 0;memset(s1, -1, sizeof s1);for (int i = 0; i < m + m; i += 2){scanf("%d%d%d", &p[i].u, &p[i].v, &p[i].w);p[i ^ 1] = p[i];swap(p[i ^ 1].v, p[i ^ 1].u);}sort(p, p + m + m);for (int i = 0; i < m + m; i++)nt1[cnt1] = s1[p[i].u], s1[p[i].u] = cnt1, e1[cnt1] = p[i].v, v1[cnt1++] = p[i].w;Dijkstra();memset(s2, -1, sizeof s2);dfs(1);work(1, n);printf("%d %d\n", ans1, ans2);}return 0;}