HDU 6181 Two Paths 次短路

题目

      HDU 6181

分析

      裸的次短路，直接套k短路模板就行了。有兴趣的读者可以看我讲k短路的博客，是利用spfa和A∗算法实现的。链接：POJ 2449 A* + spfa求第k短路 。

代码

#include <cstring>#include <cstdio>#include <queue>using namespace std;typedef long long int LL;const int INF = 0x3f3f3f3f;const int N = 100005;struct state{    LL u, g, h;    bool operator < (const state &x) const    {        return g + h > x.g + x.h;    }};struct edge{int to, cost;};vector<edge> G1[N]; // 原图vector<edge> G2[N]; // 反向图int s, t, k; // s 起点，t 终点，k 第k短路。int n, m; // n个点，m条边LL d[N];void spfa(int s){    int In_que[N] = {0};    memset(d, INF, sizeof(d));    queue<int> Q;    Q.push(s);    d[s] = 0;    In_que[s] = 1;    while (!Q.empty())    {        int u = Q.front();        Q.pop();        In_que[u] = 0;        for (unsigned i = 0; i < G2[u].size(); i++)        {            edge e = G2[u][i];            int v = e.to, dd = d[u] + e.cost;            if (d[v] > dd)            {                d[v] = dd;                if (!In_que[v])                {                    Q.push(v);                    In_que[v] = 1;                }            }        }    }}void Astar(){    int cnt[N] = {0};    priority_queue<state> Q;    Q.push({s, 0, d[s]});    while (!Q.empty())    {        state cur = Q.top();        Q.pop();        int u = cur.u;        cnt[u]++;        if (cnt[u] == k && u == t)        {            printf("%I64d\n", cur.g);            break;        }        if (cnt[u] > k)            continue;        for (unsigned i = 0; i < G1[u].size(); i++)        {            edge e = G1[u][i];            if (cnt[e.to] != k)                Q.push({e.to, cur.g + e.cost, d[e.to]});        }    }}void K_Path(int s_val, int t_val, int k_val){    s = s_val;    t = t_val;    k = k_val;    spfa(t);    Astar();}int main(){    //freopen("test.txt", "r", stdin);    //freopen("out.txt", "w", stdout);    int T;    scanf("%d", &T);    while (T--)    {        for (int i = 0; i < N; i++)        {            G1[i].clear();            G2[i].clear();        }        scanf("%d%d", &n, &m);        for (int i = 0; i < m; i++)        {            int u, v, c;            scanf("%d%d%d", &u, &v, &c);            G1[u].push_back({v, c});            G1[v].push_back({u, c});            G2[v].push_back({u, c});            G2[u].push_back({v, c});        }        K_Path(1, n, 2);    }    return 0;}