poj 2449 Remmarguts' Date （A*,k短路）

Description

给定一个有向图，求第k短路

Solution

考虑A*，f(x)=g(x)+h(x)中，g(x)表示到达当前点的代价，h(x)表示到达终点的最小可能的花费（小于等于实际花费），其中h(x)可以将边反向后SPFA得到，g(x)由搜索时递推得到

Code

#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>#include<queue>#include<vector>#include<set>#define For(i , j , k) for (int i = (j) , _##end_ = (k) ; i <= _##end_ ; ++ i)#define Fordown(i , j , k) for (int i = (j) , _##end_ = (k) ; i >= _##end_ ; -- i)#define Set(a , b) memset(a , b , sizeof(a))#define pb push_back#define mp make_pair#define x first#define y second#define INF (0x3f3f3f3f)#define Mod (1000000007)using namespace std;typedef long long LL;typedef pair<int , int> PII;template <typename T> inline bool chkmax(T &a , T b) { return a < b ? (a = b , 1) : 0; }template <typename T> inline bool chkmin(T &a , T b) { return b < a ? (a = b , 1) : 0; }int _ , __;char c_;inline int read(){    for (_ = 0 , __ = 1 , c_ = getchar() ; !isdigit(c_) ; c_ = getchar()) if (c_ == '-') __ = -1;    for ( ; isdigit(c_) ; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);    return _ * __;}inline void file(){#ifdef hany01    freopen("poj2449.in" , "r" , stdin);    freopen("poj2449.out" , "w" , stdout);#endif}const int maxn = 1010 , maxm = 100010;int n , m , s , t , k , beg[maxn] , nex[maxm] , v[maxm] , w[maxm] , dis[maxn] , beg1[maxn] , nex1[maxm] , v1[maxm] , w1[maxm] , e , e1 , times[maxn] , Ans;bool vis[maxn];queue<int> q;inline void add(int uu , int vv , int ww){    v[++ e] = vv;    w[e] = ww;    nex[e] = beg[uu];    beg[uu] = e;}inline void add1(int uu , int vv , int ww)//反向边{    v1[++ e1] = vv;    w1[e1] = ww;    nex1[e1] = beg1[uu];    beg1[uu] = e1;}inline void Init(){    int uu , vv , ww;    n = read();    m = read();    For(i , 1 , m)        uu = read(),        vv = read(),        ww = read(),        add(uu , vv , ww),        add1(vv , uu , ww);    s = read();    t = read();    k = read();    if (s == t)//坑：当起点等于终点时，最短路径长度为0，但不能计入        ++ k;}inline void SPFA()//预处理g(x){    For(i , 1 , n)        dis[i] = INF;    dis[t] = 0;    vis[t] = true;    q.push(t);    while (!q.empty())    {        int u = q.front();        q.pop();        vis[u] = false;        for (int i = beg1[u] ; i ; i = nex1[i])            if (chkmin(dis[v1[i]] , dis[u] + w1[i]))                if (!vis[v1[i]])                {                    vis[v1[i]] = true;                    q.push(v1[i]);                }    }}struct Item{    int u , g , h;    bool operator < (const Item &item) const { return g + h > item.g + item.h; }};priority_queue<Item> qi;//优先队列保存最优解inline void Astar(){    Item u;    u.u = s;    u.g = 0;    u.h = dis[s];    qi.push(u);    while (!qi.empty())    {        u = qi.top();        qi.pop();        ++ times[u.u];//记录这是第几短路        if (times[u.u] > k)//若超过k次，显然毫无意义            continue;        if (times[u.u] == k && u.u == t)//找到解，退出        {            Ans = u.h + u.g;            return ;        }        for (int i = beg[u.u] ; i ; i = nex[i])            qi.push((Item){v[i] , u.g + w[i] , dis[v[i]]});    }}int main(){    file();    Init();    SPFA();    Ans = -1;    Astar();    printf("%d\n" , Ans);    return 0;}