POJ 2449 Remmarguts' Date 第K短路 A*

题意：给出一张图，求出从s点到t点的第K短路。如果没有，输出-1.

思路：因为K<=1000,所以我们可以枚举距离从小到大所有的从s到t的路，找到第K短即可。

           这里我们用A*算法。对应的h函数是每个点到t点的最短距离，可以看出这是最终距离的下界。

          需要注意一点的是，在A*算法中，我们需要重复的加点，但是每个点至多加K次。如果加了K+1次，那对应的路的距离至少是第K+1短路。

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>

using namespace std;

int N,M,S,T,K;
const int MAXN = 1010;
const int MAXM = 100100;

struct edge{
    int to,cost;
    edge(){}
    edge(int t, int c):to(t),cost(c){}
}edges1[MAXM],edges2[MAXM];

int tot1,tot2;
int head1[MAXN],head2[MAXN];
int nxt1[MAXM],nxt2[MAXM];
int dis[MAXN];
bool used[MAXN];
int cnt[MAXN];

struct node1{
    int u, cost;
    node1(){}
    node1(int c, int uu):cost(c),u(uu){}
    bool operator < (const node1 & rhs) const{
        return cost > rhs.cost;
    }
};

struct node2{
    int u,d;
    node2(){};
    node2(int dd, int uu):u(uu),d(dd){};
    bool operator < (const node2 & rhs) const{
        return d + dis[u] > rhs.d + dis[rhs.u];
    }
};

int Astar()
{
    memset(cnt,0,sizeof(cnt));
    priority_queue<node2> Q;
    Q.push(node2(0,S));
    while(!Q.empty()){
        node2 x = Q.top();Q.pop();
        cnt[x.u]++;
        if(cnt[x.u] == K && x.u == T) return x.d;
        else if(cnt[x.u] <= K){
           for(int i = head1[x.u]; ~i; i = nxt1[i]){
                edge & e = edges1[i];
                Q.push(node2(x.d + e.cost,e.to));
            }
        }
    }
    return -1;
}

void dijstra()
{
    memset(dis,0x3f,sizeof(dis));
    memset(used,0,sizeof(used));
    dis[T] = 0;
    priority_queue<node1> Q;
    Q.push(node1(0,T));
    while(!Q.empty()){
        node1 x = Q.top();Q.pop();
        if(used[x.u]) continue;
        used[x.u] = true;
        for(int i = head2[x.u]; ~i; i = nxt2[i]){
            edge & e = edges2[i];
            if(dis[e.to] > x.cost + e.cost){
                dis[e.to] = x.cost + e.cost;
                Q.push(node1(dis[e.to],e.to));
            }
        }
    }
}

void init()
{
    tot1 = tot2 = 0;
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
}
void addedge(int u, int v, int c)
{
    edges1[tot1] = edge(v,c);
    nxt1[tot1] = head1[u];
    head1[u] = tot1++;

    edges2[tot2] = edge(u,c);
    nxt2[tot2] = head2[v];
    head2[v] = tot2++;
}

int main(void)
{
    //freopen("input.txt","r",stdin);
    scanf("%d%d", &N, &M);
    init();
    for(int i = 0; i < M; ++i){
        int a,b,t;
        scanf("%d%d%d",&a,&b,&t);
        addedge(a,b,t);
    }
    scanf("%d%d%d",&S,&T,&K);
    if(S == T) K++;
    dijstra();
    printf("%d\n",Astar());
    return 0;
}