【POJ2449】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;
}