题解:给一个1e5个点2e5条边,每个边有一个值,让你输出一条从1到n边的路径使得:条数最短的前提下字典序最小。
题解:bfs一次找最短路(因为权值都是1,不用dijkstra),再bfs一次存一下路径,4个月前的代码(忘了为什么搞得那么麻烦),wa了两天,今天看了一下题目,看了一下代码,改了一下初始化数组,直接过了orz
#define _CRT_SECURE_NO_WARNINGS
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdio>
#include<string>
#include<stack>
#include<list>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<sstream>
#include<iostream>
#include<algorithm>
//std::ios::sync_with_stdio(false);
using namespace std;
const int maxn = 2e5 + ;
vector<pair<int, int> >E[maxn];
int d[maxn];
int n, m;
void bfs(int n) {
queue<int> Q;
Q.push(n); d[n] = ;
while (!Q.empty()) {
int now = Q.front(); Q.pop();
for (int i = ; i < E[now].size(); i++) {
int v = E[now][i].first;
if (d[v])continue;
d[v] = d[now] + ;
Q.push(v);
}
}
}
int ans[maxn];
int vis[maxn];
void bfs1() {
for (int i = ; i <= n; i++)ans[i] = 1e9+;
queue<int>Q;
Q.push();
while (!Q.empty()) { int now = Q.front(); Q.pop();
if (vis[now])continue;
vis[now] = ;
int mn = 1e9 + ;
for (int i = ; i < E[now].size(); i++) {
int v = E[now][i].first;
if (d[v] != d[now] - )continue;
int w = E[now][i].second;
mn = min(mn, w);
}
int diff = d[] - d[now];
ans[diff] = min(ans[diff], mn);
for (int i = ; i < E[now].size(); i++) {
int v = E[now][i].first;
if (d[v] != d[now] - )continue;
int w = E[now][i].second;
if (w == mn) Q.push(v); } }
}
void init() {
memset(ans, , sizeof(ans));
memset(vis, , sizeof(vis));
memset(d, , sizeof(d));
for (int i = ; i <= n; i++)E[i].clear();
}
int main() {
while (cin >> n >> m) {
init();
for (int i = ; i <= m; i++) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
E[x].push_back(make_pair(y, z));
E[y].push_back(make_pair(x, z));
}
bfs(n);
bfs1();
cout << d[] - << endl;
for (int i = ; i < d[] - ; i++)printf("%d%c", ans[i], i + == d[] - ? '\n' : ' ');// << ' ';| }
//system("pause");
return ;
}
/* 4 6
1 2 1
1 3 2
3 4 3
2 3 1
2 4 4
3 1 1 */