---
title: Recover Path
tags:
  - Dynamic Programming
  - Shortest Path
id: "3154"
categories:
  - [OI, Common Skill]
  - [OI, Graph Theory]
date: 2018-05-17 08:35:49
---

## 题目描述

Traveller Gregory is famous for his ability to always choose the shortest path for his journey. Ostap is a journalist who seeks for information about the recent Gregory's trip. He managed to get the evidence that during this trip Gregory visited a number of cities. However, there is no information about the order in which they were visited, and no information about the starting city and the ending city of Gregory's trip (the entire trip is (one of) the shortest path between these cities). Help Ostap to find any shortest path that contains all specified cities.

Country in which Gregory traveled consists of $n$ cities and $m$ undirected roads between them. For each road Ostap knows the time it takes to travel it, and the "shortest" word above is with respect to those times.

It is guaranteed that there exists some shortest path going through all specified cities.

## 题意概述

给定一张$n$个点$m$条边的无向图，第$i$条边的权值为$t_i$。给定其中$k$个点，要求找一对点$s, t$，使得这$k$个点都在从$s$到$t$的某条最短路径上，并输出这条路径。保证有解。

数据范围：$1 \le n, m \le 10^5, \; 1 \le t_i \le 10000$。

## 算法分析

显然，存在一种方案使得$s, t$在给定的$k$个点中。

从$k$个点中任选一个点，找到在$k$个点中距离它最远的点$u$，这个点一定是$s$或$t$，可以用反证法证明。接着从$u$开始跑最短路，得到一张拓扑图。令$f_i$表示从$i$到$u$的最短路径上最多包含多少个$k$个点中的点。只要按距离从小到大枚举$i$，再枚举它的连边是否可以作为最短路径中的边。同时还要记录方案。

## 代码

```cpp
#include <algorithm>
#include <cstdio>
#include <cstring>

static int const N = 100005;
int ne = 1, top, h[N], d[N], que[N], id[N], lst[N], pre[N], ans[N];
bool in[N], v[N];
struct Edge {
  int u, v, w, nxt;
} e[N << 1];

void add_edge(int u, int v, int w) {
  e[++ne] = (Edge){u, v, w, h[u]}, h[u] = ne;
  e[++ne] = (Edge){v, u, w, h[v]}, h[v] = ne;
}

void spfa(int s) {
  int qb = 0, qe = 1;
  memset(d, 0x3f, sizeof d), d[s] = 0, que[0] = s, in[s] = 1;
  for (; qb != qe;) {
    int u = que[qb++];
    in[u] = 0, qb == N && (qb = 0);
    for (int i = h[u]; i; i = e[i].nxt)
      if (d[e[i].v] > d[u] + e[i].w) {
        d[e[i].v] = d[u] + e[i].w;
        if (!in[e[i].v]) {
          if (qb == qe || d[e[i].v] > d[que[qb]])
            que[qe++] = e[i].v, in[e[i].v] = 1, qe == N && (qe = 0);
          else
            !~--qb && (qb = N - 1), que[qb] = e[i].v, in[e[i].v] = 1;
        }
      }
  }
}

int main() {
  int n, m, k, t;
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v, w; i <= m; ++i)
    scanf("%d%d%d", &u, &v, &w), add_edge(u, v, w);
  scanf("%d%d", &k, &t), v[t] = 1, spfa(t);
  int s = t;
  for (int i = 2; i <= k; ++i)
    scanf("%d", &t), v[t] = 1, d[s] < d[t] && (s = t);
  spfa(s);
  for (int i = 1; i <= n; ++i)
    id[i] = i;
  std::sort(id + 1, id + n + 1, [&](int const &i, int const &j) { return d[i] < d[j]; });
  for (int i : id) {
    lst[i] += v[i];
    for (int j = h[i]; j; j = e[j].nxt)
      if (d[e[j].v] == d[i] + e[j].w && lst[e[j].v] < lst[i])
        lst[e[j].v] = lst[i], pre[e[j].v] = j;
  }
  for (int i = 1; i <= n; ++i)
    if (lst[i] == k) {
      for (int j = pre[i]; j; j = pre[e[j].u])
        ans[top++] = j >> 1;
      break;
    }
  printf("%d\n", top);
  for (int i = 0; i < top; ++i)
    printf("%d ", ans[i]);
  puts("");
  return 0;
}
```
