HDU-2017 多校训练赛10-1011-Two Paths

ACM模版

描述

题解

这个题是裸的次短路算法，用 A∗ 算法比较好一些。

代码

#include <iostream>#include <cstdio>#include <queue>#include <cstring>using namespace std;typedef long long ll;const ll INF = 1e18;const int MAXN = 1e5 + 5;int n, m;int tot, st, ed, k;ll dis[MAXN];int hed1[MAXN];int hed2[MAXN];bool vis[MAXN];struct Edge{    int u, v, nxt1, nxt2;    ll c;    Edge() {}    Edge(int _u, int _v, ll _c) : u(_u), v(_v), c(_c) {}} e[MAXN << 1];struct qnode{    int v;    ll c;    qnode() {}    qnode(int _v, ll _c) : v(_v), c(_c) {}    bool operator < (const qnode &rhs) const    {        return c + dis[v] > rhs.c + dis[rhs.v];    }};void addedge(int u, int v, ll c){    e[tot] = Edge(u, v, c);    e[tot].nxt1 = hed1[u];    hed1[u] = tot;    e[tot].nxt2 = hed2[v];    hed2[v] = tot++;}void dij(int src){    memset(vis, false, sizeof(vis));    for (int i = 1; i <= n; i++)    {        dis[i] = INF;    }    dis[src] = 0;    priority_queue<qnode> que;    que.push(qnode(src, 0));    while (!que.empty())    {        qnode pre = que.top();        que.pop();        if (vis[pre.v])        {            continue;        }        vis[pre.v] = true;        for (int i = hed2[pre.v]; i != -1; i = e[i].nxt2)        {            if (dis[e[i].u] > dis[pre.v] + e[i].c)            {                dis[e[i].u] = dis[pre.v] + e[i].c;                que.push(qnode(e[i].u, 0));            }        }    }}ll a_star(int src){    priority_queue<qnode> que;    que.push(qnode(src, 0));    k--;    while (!que.empty())    {        qnode pre = que.top();        que.pop();        if (pre.v == ed)        {            if (k)            {                k--;            }            else            {                return pre.c;            }        }        for (int i = hed1[pre.v]; i != -1; i = e[i].nxt1)        {            que.push(qnode(e[i].v, pre.c + e[i].c));        }    }    return -1;}int main(){    int T;    scanf("%d", &T);    while (T--)    {        tot = 0;        memset(hed1, -1, sizeof(hed1));        memset(hed2, -1, sizeof(hed2));        scanf("%d%d", &n, &m);        int u, v;        ll c;        for (int i = 0; i < m; i++)        {            scanf("%d%d%lld", &u, &v, &c);            addedge(u, v, c);            addedge(v, u, c);        }        st = 1, ed = n, k = 2;        dij(ed);        if (st == ed)        {            k++;        }        printf("%lld\n", a_star(st));    }    return 0;}