545E. Paths and Trees

题目链接

题意：给定一个无向图和一个点u，找出若干条边组成一个子图，要求这个子图中u到其他个点的最短距离与在原图中的相等，并且要求子图所有边的权重之和最小，求出最小值并输出子图的边号。

思路：先求一遍最短路，从所有到i点的满足最短路的边中选一条权最小的边。

Java程序

import java.io.PrintStream;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class E545 {
private static class Edge {
int v;
long w;
int index;

Edge(int v, long w, int index) {
this.v = v;
this.w = w;
this.index = index;
}
}

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
PrintStream out = System.out;

int n = in.nextInt(), m = in.nextInt();
List<Edge>[] graph = new List
;

for (int i = 0; i < n; i++) {
graph[i] = new ArrayList<E545.Edge>();
}

for (int i = 1; i <= m; i++) {
int v1 = in.nextInt() - 1;
int v2 = in.nextInt() - 1;
long w = in.nextLong();

graph[v1].add(new Edge(v2, w, i));
graph[v2].add(new Edge(v1, w, i));
}
int u = in.nextInt() - 1;

Edge[] lastEdge = new Edge
;
final long[] min = new long
;
for (int i = 0; i < n; i++) {
min[i] = -1;
}

min[u] = 0;
Queue<Integer> q = new LinkedList<Integer>();

q.add(u);

while (!q.isEmpty()) {
int v = q.poll();

for (Edge edge : graph[v]) {
int v1 = edge.v;
long min1 = min[v] + edge.w;

if ((min[v1] == -1) || (min1 < min[v1])
|| (min1 == min[v1] && edge.w < lastEdge[v1].w)) {

min[v1] = min1;
lastEdge[v1] = edge;
q.add(v1);
}
}
}

long res = 0;
boolean[] f = new boolean[m];

for (int i = 0; i < n; i++) {
if (lastEdge[i] != null) {
res += lastEdge[i].w;
f[lastEdge[i].index - 1] = true;
}
}

out.println(res);

StringBuilder s = new StringBuilder();
boolean first = true;
for (int i = 0; i < m; i++) {
if (f[i]) {
if (!first) {
s.append(" ");
}
s.append(i + 1);
first = false;
}
}
out.println(s.toString());
in.close();
out.close();

}

}

 

Python代码

import heapq as hq

class edge(object):
def __init__(self, to, w, nr):
self.to = to
self.w = w
self.nr = nr

n, m = map(int, raw_input().split())
adj = [[] for _ in range(n + 1)]
for i in range(1, m+1):
u, v, c = map(int, raw_input().split())
adj[u].append((v, c, i))
adj[v].append((u, c, i))
root = int(raw_input())
vis = [False] * (n+1)
q = [(0, 0, root, 0)]
ans = []
tot = 0
while q:
d, c, n, e = hq.heappop(q)
if vis
:
continue
ans.append(e)
tot += c
vis
= True
for v, c, i in adj
:
if not vis[v]:
hq.heappush(q, (d+c, c, v, i))
ans = map(str, ans)
print tot
print " ".join(ans[1:])

 

上面的代码都是在codeforces上面抄过来的，自己写不出来。。。。