UVA 1599 Ideal Path

题目链接：https://vjudge.net/problem/UVA-1599

题目分析与翻译摘自《算法禁赛入门经典》

题目大意

　　给一个 n 个点 m 条边（2 ≤ n ≤ 100000，1 ≤ m ≤ 200000）的无向图，每条边上都涂有一种颜 色。求从结点 1 到结点 n 的一条路径，使得经过的边数尽量少，在此前提下，经过边的颜色序列的字典序最小。一对结点间可能有多条边，一条边可能连接两个相同结点。输入保证结点 1 可以达到结点 n。颜色为 1～10⁹ 的整数。

分析

　　从终点开始“倒着”BFS，得到每个结点 i 到终点的最短距离 d[i]，然后直接从起点开始走，但是每次到达一个新结点时要保证 d 值恰好减少 1（如有多个选择则可以随便走），直到到达终点。可以证明：这样走过的路径一定是一条最短路。

　　从起点开始按照上述规则走，如果有多种走法，选颜色字典序最小的走；如果有多条边的颜色字典序都是最小，则记录所有这些边的终点，走下一步时要考虑从所有这些点出发的边（具体实现就是边记录最小值边放入队列，然后在出队的时候判断）。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x7f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
    Rep(i, v.size()) out << v[i] << " 
"[i == v.size()];
    return out;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< double > VD;
typedef vector< VI > VVI;
typedef vector< SI > VSI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< int, string > MIS;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef map< string, string > MSS;
typedef map< PII, string > MPIIS;
typedef map< PII, PII > MPIIPII;
typedef multimap< int, int > MMII;
typedef multimap< string, int > MMSI;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct Edge{
    int from, to, c;
};

int m, n, k;
vector< Edge > e;
VI v[maxN];
// path[i] 表示第 i 条边能取到的颜色的最小值 
int d[maxN], path[maxN];
bool vis[maxN];

inline void addEdge(Edge &x) {
    v[x.from].PB(e.size());
    e.PB(x);
    swap(x.from, x.to);
    v[x.from].PB(e.size());
    e.PB(x);
}

inline void bfs1() {
    queue< int > Q;
    d[n] = 0;
    ms0(vis);
    vis[n] = 1;
    Q.push(n);
    
    while(!Q.empty()) {
        int tmp = Q.front(); Q.pop();
        
        foreach(i, v[tmp]) {
            Edge &x = e[*i];
            if(vis[x.to]) continue;
            vis[x.to] = 1;
            
            d[x.to] = d[tmp] + 1;
            
            if(x.to == 1) return;
            Q.push(x.to);
        }
    }
}

inline void bfs2() {
    queue< PII > Q;
    // memset 是以8个bit为单位初始化的！！！ 
    msI(path);
    path[0] = 0;
    ms0(vis);
    Q.push(PII(1, 0));
    
    while(!Q.empty()) {
        // tmpv 为当前顶点， tmpc为到达当前顶点的边的颜色 
        int tmpv = Q.front().ft, tmpc = Q.front().sd; Q.pop();
        
        if(path[k - d[tmpv]] < tmpc || vis[tmpv]) continue;
        vis[tmpv] = 1; // 到达一个节点的路径可能有多条，不记录的话会被重复访问，复杂度指数上升 
        
        foreach(i, v[tmpv]) {
            Edge &x = e[*i];
            if(d[tmpv] - 1 != d[x.to] || d[x.to] == 0 && x.to != n) continue; 
            path[k - d[x.to]] = min(path[k - d[x.to]], x.c);
            
            Q.push(PII(x.to, x.c));
        }
    }
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    while(~scanf("%d%d", &n, &m)) {
        e.clear();
        For(i, 1, n) v[i].clear(); 
        
        Rep(i, m) {
            Edge t;
            scanf("%d%d%d", &t.from, &t.to, &t.c);
            addEdge(t);
        }
        
        bfs1();
        k = d[1];
        
        bfs2();
        
        printf("%d
", k);
        For(i, 1, k) printf("%d%c", path[i], " 
"[i == k]);
    }
    return 0;
}