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)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;
}