【图论】【启发式搜索】[POJ 2449]Remmarguts' Date

实际上就是从T到当前节点的最短路的长度，因为首先要保证h(n) <= h*(n)且h(n)尽量接近h*(n)那么直接令h(n)=h*(n)就行了，然后注意每一次判断是否到达终点的时候要在每一次取出的时候判断，因为这个时候的最短路是经过优先队列的排序的。还有就是注意下S==T的情况默认的时候S到本身的最短路肯定是0，那么我们如果要求现在的最短路那么求得就是当前的次短路，让k++就好了

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int MAXN = 1000;
const int MAXM = 100000;
struct node{
    int v, w;
    node *next;
}Edges[MAXM*2+10], *ecnt = Edges, *adj[MAXN+10], *adj1[MAXN+10];
struct State{
    int f, id;
    bool operator<(const State& s) const { return f>s.f; }
};
priority_queue<State> que;
void addedge(int u, int v, int w){
    ++ecnt;
    ecnt->v = v;
    ecnt->w = w;
    ecnt->next = adj[u];
    adj[u] = ecnt;
}
void _addedge(int u, int v, int w){
    ++ecnt;
    ecnt->v = v;
    ecnt->w = w;
    ecnt->next = adj1[u];
    adj1[u] = ecnt;
}
int dis[MAXN+10];
bool inque[MAXN+10];
void SPFA(int res){
    queue<int> que;
    que.push(res); inque[res] = true;
    memset(dis, 0x3f, sizeof dis);
    dis[res] = 0;
    while(!que.empty()){
        res = que.front(); que.pop(); inque[res] = false;
        for(node *p=adj1[res];p;p=p->next){
            int v=p->v;
            if(dis[v] > dis[res] + p->w){
                dis[v] = dis[res] + p->w;
                if(!inque[v]){
                    inque[v] = true;
                    que.push(v);
                }
            }
        }
    }
}
int main(){
    int n, m;
    int S, T, K, u, v, w;
    scanf("%d%d", &n, &m);
    for(int i=1;i<=m;i++){
        scanf("%d%d%d", &u, &v, &w);
        addedge(u, v, w);
        _addedge(v, u, w);
    }
    scanf("%d%d%d", &S, &T, &K);
    if(S == T) K++;
    SPFA(T);
    State t1, t2;
    t1.f = dis[S]; t1.id = S;
    que.push(t1);
    int counter = 0;
    while(!que.empty()){
        t1 = que.top(); que.pop();
        if(t1.id == T){
            counter++;
            if(counter == K){
                printf("%d\n" ,t1.f);
                return 0;
            }
        }
        for(node *p=adj[t1.id];p;p=p->next){
            int v = p->v;
            t2.f = t1.f - dis[t1.id] + p->w + dis[v];
            t2.id = v;
            que.push(t2);
        }
    }
    printf("-1\n");

    return 0;
}