每日一题 UVA - 1599 Ideal Path 字典序最短路

Posted 2021-01-02 suut

 

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 + 5;
vector<pair<int, int> >E[maxn];
int d[maxn];
int n, m;
void bfs(int n) {
    queue<int> Q;
    Q.push(n); d[n] = 1;
    while (!Q.empty()) {
        int now = Q.front(); Q.pop();
        for (int i = 0; i < E[now].size(); i++) {
            int v = E[now][i].first;
            if (d[v])continue;
            d[v] = d[now] + 1;
            Q.push(v);
        }
    }
}
int ans[maxn];
int vis[maxn];
void bfs1() {
    for (int i = 0; i <= n; i++)ans[i] = 1e9+5;
    queue<int>Q;
    Q.push(1);
    while (!Q.empty()) {

        int now = Q.front(); Q.pop();
        if (vis[now])continue;
        vis[now] = 1;
        int mn = 1e9 + 5;
        for (int i = 0; i < E[now].size(); i++) {
            int v = E[now][i].first;
            if (d[v] != d[now] - 1)continue;
            int w = E[now][i].second;
            mn = min(mn, w);
        }
        int diff = d[1] - d[now];
        ans[diff] = min(ans[diff], mn);
        for (int i = 0; i < E[now].size(); i++) {
            int v = E[now][i].first;
            if (d[v] != d[now] - 1)continue;
            int w = E[now][i].second;
            if (w == mn) Q.push(v);

        }

    }
}
void init() {
    memset(ans, 0, sizeof(ans));
    memset(vis, 0, sizeof(vis));
    memset(d, 0, sizeof(d));
    for (int i = 0; i <= n; i++)E[i].clear();
}
int main() {
    while (cin >> n >> m) {
        init();
        for (int i = 1; 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[1] - 1 << endl;
        for (int i = 0; i < d[1] - 1; i++)printf("%d%c", ans[i], i + 1 == d[1] - 1 ? ‘
‘ : ‘ ‘);// << ‘ ‘;|

    }
    //system("pause");
    return 0;
}
/*

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

*/