hdu Two Paths 次短路模板 (可往回走)

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#define MAXN (200000 + 10)#define INF 0x3f3f3f3fcusing namespace std;typedef long long LL;struct edge{    int to, cost;    edge(int tv = 0, int tc = 0):        to(tv), cost(tc) {}};typedef pair<LL,LL> P;int N, R;vector<edge> graph[MAXN];LL dist[MAXN];     //最短距离LL dist2[MAXN];    //次短距离void solve(){    fill(dist, dist+N, INF);    fill(dist2, dist2+N, INF);    //从小到大的优先队列    //使用pair而不用edge结构体    //是因为这样我们不需要重载运算符    //pair是以first为主关键字进行排序    priority_queue<P, vector<P>, greater<P> > Q;    //初始化源点信息    dist[0] = 0;    Q.push(P(0, 0));    //同时求解最短路和次短路    while(!Q.empty())    {        P p = Q.top();        Q.pop();        //first为s->to的距离，second为edge结构体的to        LL v = p.second, d = p.first;        //当取出的值不是当前最短距离或次短距离，就舍弃他        if(dist2[v] < d) continue;        for(unsigned i = 0; i < graph[v].size(); i++)        {            edge &e = graph[v][i];            LL d2 = d + e.cost;            if(dist[e.to] > d2)            {                swap(dist[e.to], d2);                Q.push(P(dist[e.to], e.to));            }            if(dist2[e.to] > d2 && dist[e.to] < d2)            {                dist2[e.to] = d2;                Q.push(P(dist2[e.to], e.to));            }        }    }    printf("%lld\n", dist2[N-1]);}int main(){    int A, B, D;    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &N, &R);        for(int i=0;i<N;++i)            graph[i].clear();        for(int i = 0; i < R; i++)        {            scanf("%d%d%d", &A, &B, &D);            graph[A-1].push_back(edge(B-1, D));            graph[B-1].push_back(edge(A-1, D));        }        solve();    }    return 0;}

标程如下

#include <bits/stdc++.h>#define mp make_pairusing namespace std;typedef long long LL;typedef pair<int, int> PII;typedef pair<LL, int> PLI;struct Edge {    int from, to;    Edge( int f = 0, int t = 0 ) : from(f), to(t) {        if( f > t ) swap( f, t );    }};bool operator < ( const Edge& a, const Edge& b ) {    if( a.from != b.from )  return a.from < b.from;    return a.to < b.to;}void shortestPath( int from, LL d[], int prv[], vector<PII> g[], int n ) {    priority_queue<PLI, vector<PLI>, greater<PLI>> pq;    while( !pq.empty() )    pq.pop();    for(int i = 1; i <= n; i++ )    d[i] = -1;    d[from] = 0;    prv[from] = from;    pq.push( mp( d[from], from ) );    while ( !pq.empty() ) {        PLI now = pq.top();        pq.pop();        for( PII nxt : g[now.second] ) {            int v = nxt.first;            int w = nxt.second;            if( d[v] == -1 ) {                d[v] = now.first + w;                pq.push( mp( d[v], v ) );                prv[v] = now.second;                continue;            }            if( d[v] > now.first + w ) {                d[v] = now.first + w;                pq.push( mp( d[v], v ) );                prv[v] = now.second;            }        }    }}const int MAXN = 100006;vector<PII> g[MAXN];int prv[MAXN];LL d1[MAXN], d2[MAXN];set<Edge> used;set<Edge> chk;vector<int> usedw;void addEdge( int& u, int& v, int& w ) {    g[u].emplace_back( v, w );    g[v].emplace_back( u, w );}int main() {    //freopen("paths.in", "r", stdin );    //freopen("paths.out", "w", stdout );    int n, m;    int T;    scanf("%d", &T);    while(T--) {        assert(scanf("%d%d", &n, &m) == 2);        chk.clear();        assert( 1 <= n and n <= 100000 );        assert( 1 <= m and m <= 100000 );        for(int i = 1; i <= n; i++ )    g[i].clear();        for(int i = 1; i <= m; i++ ) {            int u, v, w;            assert( scanf("%d%d%d", &u, &v, &w ) == 3 );            addEdge( u, v, w );            assert( u >= 1 and u <= n );            assert( v >= 1 and v <= n );            assert( w >= 1 and w <= 1000000000 );            assert( u != v );            assert( chk.count( Edge( u, v ) ) == 0 );            chk.insert( Edge( u, v ) );        }        shortestPath( n, d2, prv, g, n );        shortestPath( 1, d1, prv, g, n );        int to = n;        used.clear();        usedw.clear();        while( to != 1) {            int from = prv[to];            used.insert( Edge( from, to ) );            usedw.push_back(d1[to] - d1[from]);            to = from;        }        LL ans = (1LL << 60 );        for(int u = 1; u <= n; u++ ) {            for( PII dst : g[u] ) {                int v = dst.first;                if( !used.count( Edge( u, v ) ) ) {                    if( ~d1[u] and ~d2[v] ) {                        ans = min( ans, d1[u] + d2[v] + dst.second );                    }                }            }        }        for (auto item : usedw) {            LL tmp = d1[n] + item * 2;            ans = min(ans, tmp);        }        if( ans == (1LL << 60 ) )   assert(0);        else    printf("%lld\n", ans );    }    return 0;}