CF20C Dijkstra? 最短路

题意翻译

题目大意

给出一张图，请输出其中任意一条可行的从点 111 到点 nnn 的最短路径。

输入输出格式

输入格式

第一行：两个整数n，m，分别表示点数和边数

接下来m行：每行三个整数u，v，w，表示u和v之间连一条边权为w的双向边。

输出格式

一行：一个可行的路径，如果不存在这种路径输出-1

2<=n<=10^5，0<=m<=105

题目描述

You are given a weighted undirected graph. The vertices are enumerated from 1 to n n n . Your task is to find the shortest path between the vertex 1 1 1 and the vertex n n n .

输入格式

The first line contains two integers n n n and m m m ( 2<=n<=105,0<=m<=105 2<=n<=10^{5},0<=m<=10 2<=n<=105,0<=m<=105 ), where n n n is the number of vertices and m m m is the number of edges. Following m m m lines contain one edge each in form ai a_{i} ai , bi b_{i} bi and wi w_{i} wi ( 1<=ai,bi<=n,1<=wi<=106 1<=a_{i},b_{i}<=n,1<=w_{i}<=10^{6} 1<=ai,bi<=n,1<=wi<=106 ), where ai,bi a_{i},b_{i} ai,bi are edge endpoints and wi w_{i} wi is the length of the edge.

It is possible that the graph has loops and multiple edges between pair of vertices.

输出格式

Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.

输入输出样例

输入 #1

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

输出 #1

1 4 3 5

输入 #2

5 6
1 2 2
2 5 5
2 3 4
1 4 1
4 3 3
3 5 1

输出 #2

1 4 3 5

题解

最短路模板题。。。

在计算最短路过程中，pre[i]记录点 i 的最短路前驱。

利用pre[i]数组输出答案即可。

注意最后一个点范围爆long long.

code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 2e5 + 5;
int n, m;
struct edge { int to, nxt, val; } e[N];
int cnt, head[N];
void add(int from, int to, int val) {
	e[++ cnt].to = to;
	e[cnt].val = val;
	e[cnt].nxt = head[from];
	head[from] = cnt;
}
int dis[N], vis[N], pre[N];
LL ans[N];
void spfa() {
	for(int i = 1;i <= n;i ++) dis[i] = 1e15;
	queue <int> q; q.push(1);
	dis[1] = 0; vis[1] = 1;
	while(! q.empty()) {
		int tp = q.front(); q.pop();
		vis[tp] = 0;
		for(int i = head[tp]; i ;i = e[i].nxt) {
			int to = e[i].to, val = e[i].val;
			if(dis[to] > dis[tp] + val) {
				dis[to] = dis[tp] + val;
				pre[to] = tp;
				if(! vis[to]) q.push(to), vis[to] = 1;
			}
		}
	}
}
signed main() {
	cin >> n >> m;
	for(int i = 1, a, b, l;i <= m;i ++) {
		cin >> a >> b >> l;
		add(a, b, l); add(b, a, l);
	}
	spfa();
	if(dis[n] == 1e15) { cout << "-1"; return 0; }
	int t = n, tot = 0;
	while(t != 1) {
		ans[++ tot] = t;
		t = pre[t];
	}
	ans[++ tot] = 1;
	for(int i = tot; i ;i --) cout << ans[i] << " ";
	return 0;
}

就这样，我又水了一篇题解。