POJ 2449 Remmarguts' Date

Posted 2020-12-02 evenbao

【题目链接】

            http://poj.org/problem?id=2449

【算法】

            A*（启发式搜索）

            首先，求第k短路可以用优先队列BFS实现，当T第k次入队时，就求得了第k短路，但是，这种做法的复杂度太高

            考虑使用A*算法，每个点的估价函数就是这个点到T的最短路，不妨将所有的边反过来求最短路，这样就得到了所有点的估价函数

            这种算法的复杂度是优秀的

  【代码】

           

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 1010
#define MAXM 100010
const int INF = 2e9;

int i,n,m,tot,rtot,u,v,w,S,T,K;
int head[MAXN],rhead[MAXN],dist[MAXN];

struct Edge
{
        int to,w,nxt;
} e[MAXM<<1];
struct info
{
        int s,d;
        friend bool operator < (info a,info b)
        {
                return a.d + dist[a.s] > b.d + dist[b.s]; 
        }        
} ;

inline void add(int u,int v,int w)
{
        tot++;
        e[tot] = (Edge){v,w,head[u]};
        head[u] = tot;
        tot++;
        e[tot] = (Edge){u,w,rhead[v]};
        rhead[v] = tot;        
}
inline void dijkstra(int s)
{
        int i,v,w;
        priority_queue< pair<int,int> > q;
        pair<int,int> cur;
        static bool visited[MAXN];
        while (!q.empty()) q.pop();
        for (i = 1; i <= n; i++) 
        {
                dist[i] = INF;
                visited[i] = false;
        }
        dist[s] = 0;
        q.push(make_pair(0,s));
        while (!q.empty())
        {
                cur = q.top();
                q.pop();
                if (visited[cur.second]) continue;
                visited[cur.second] = true;
                for (i = rhead[cur.second]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        if (dist[cur.second] + w < dist[v])
                        {
                                dist[v] = dist[cur.second] + w;
                                q.push(make_pair(-dist[v],v));
                        }
                }
        }         
}
inline int Astar(int s,int t,int k)
{
        int i,cnt = 0,v,w;
        priority_queue< info > q;
        info cur;
        while (!q.empty()) q.pop();        
        q.push((info){s,0});
        while (!q.empty())
        {
                cur = q.top();
                q.pop();
                if (cur.s == t)
                {
                        cnt++;
                        if (cnt == k) return cur.d;
                }
                for (i = head[cur.s]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        q.push((info){v,cur.d+w});
                }
        }
        return -1;
}

int main() 
{
        
        while (scanf("%d%d",&n,&m) != EOF)
        {
                tot = rtot = 0;
                for (i = 1; i <= n; i++) head[i] = rhead[i] = 0;
                for (i = 1; i <= m; i++)
                {
                        scanf("%d%d%d",&u,&v,&w);
                        add(u,v,w);
                }
                scanf("%d%d%d",&S,&T,&K);
                if (S == T) K++;
                dijkstra(T);
                printf("%d
",Astar(S,T,K));
        }
        
        return 0;
    
}