CF 545E Paths and Trees

题目描述

http://codeforces.com/problemset/problem/545/E

简要题意:给定一个 $n$ 个点 $m$ 条正权边的无向图和一个源点 $s$,求边权和最小的最短路径树

$n,m\le 3\times 10^5$

Solution

需要注意的是我们不能将最短路径图求出来之后跑 $kruskal$,最短路径图的边是有向的

我们直接贪心对于每个点拿最短的边即可,注意这样一定有解

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#define maxn 300010
#define ll long long
#define INF 1000000000000000000ll
using namespace std;

int n, m, s;

struct Edge {
int fr, to, next, w;
} e[maxn * 2]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].fr = u; e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

struct Queue {
int k; ll d;

friend bool operator < (const Queue &u, const Queue &v) { return u.d > v.d; }
}; bool vis[maxn]; ll dis[maxn]; int pre[maxn];
priority_queue<Queue> Q;
void dijkstra() {
fill(dis, dis + maxn, INF); dis[s] = 0;
Q.push({ s, 0 });
while (!Q.empty()) {
int u = Q.top().k; Q.pop();
if (vis[u]) continue; vis[u] = 1;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
pre[v] = i;
Q.push({ v, dis[v] });
}
else if (dis[v] == dis[u] + w && e[pre[v]].w > w) pre[v] = i;
}
}
}

int main() { fill(head, head + maxn, -1);
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m;
for (int i = 1; i <= m; ++i) {
int x, y, z; cin >> x >> y >> z;
add_edge(x, y, z); add_edge(y, x, z);
} cin >> s; dijkstra();
ll ans = 0;
for (int i = 1; i <= n; ++i) if (i != s) ans += e[pre[i]].w;
cout << ans << "\n";
for (int i = 1; i <= n; ++i)
if (i != s) cout << pre[i] / 2 + 1 << " \n"[i == n];
return 0;
}