POJ 2449 Remmarguts' Date (算竞进阶习题)

A* + dijkstra/spfa

第K短路的模板题，就是直接把最短路当成估价函数，保证估价函数的性质(从当前状态转移的估计值一定不大于实际值)
我们建反图从终点跑最短路，就能求出从各个点到终点的最短距离，这样就能满足估价函数的性质了
要注意一点，当起点和终点一样的时候第k短路就变成k+1短了，因为0也算一条。。。

话说回来为啥我用pair就MLE了呢。。。。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C yql){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
    return ans;
}
const int N = 1005;
int n, m, k, s, t, cnt1, cnt2, head1[N], head2[N], dist[N], num[N], w[N];
bool vis[N];
struct Edge{ int v, next, dis; } edge1[100005], edge2[100005];

struct Point{
    int s, dis;
    bool operator < (const Point &rhs) const {
        return dis > rhs.dis;
    }
    Point(int s, int dis): s(s), dis(dis){}
};

struct Node{
    int v, f, g;
    Node(int v, int f, int g): v(v), f(f), g(g){}
    bool operator < (const Node &rhs) const{
        return f + g > rhs.f + rhs.g;
    }
};

void addEdge1(int a, int b, int c){
    edge1[cnt1].v = b, edge1[cnt1].next = head1[a], edge1[cnt1].dis = c;
    head1[a] = cnt1 ++;
}

void addEdge2(int a, int b, int c){
    edge2[cnt2].v = b, edge2[cnt2].next = head2[a], edge2[cnt2].dis = c;
    head2[a] = cnt2 ++;
}

void dijkstra(){
    fill(dist, dist + N, INF);
    priority_queue<Point> pq;
    dist[t] = 0;
    pq.push(Point(t, dist[t]));
    while(!pq.empty()){
        int s = pq.top().s, d = pq.top().dis; pq.pop();
        if(vis[s]) continue;
        vis[s] = true;
        for(int i = head2[s]; i != -1; i = edge2[i].next){
            int u = edge2[i].v;
            if(dist[u] > d + edge2[i].dis){
                dist[u] = d + edge2[i].dis;
                pq.push(Point(u, dist[u]));
            }
        }
    }
}

int astar(){
    if(dist[s] == INF) return -1;
    if(s == t) k ++;
    priority_queue<Node> pq;
    pq.push(Node(s, dist[s], 0));
    while(!pq.empty()){
        Node cur = pq.top(); pq.pop();
        num[cur.v] ++;
        if(cur.v == t && num[cur.v] == k) return cur.f + cur.g;
        if(num[cur.v] > k) continue;
        for(int i = head1[cur.v]; i != -1; i = edge1[i].next){
            int u = edge1[i].v;
            if(num[u] == k) continue;
            pq.push(Node(u, dist[u], cur.g + edge1[i].dis));
        }
    }
    return -1;
}

void init(){
    cnt1 = cnt2 = 0;
    memset(head1, -1, sizeof head1);
    memset(head2, -1, sizeof head2);
}

int main(){

    while(scanf("%d%d", &n, &m) != EOF){
        init();
        for(int i = 0; i < m; i++){
            int a, b, c; scanf("%d%d%d", &a, &b, &c);
            addEdge1(a, b, c), addEdge2(b, a, c);
        }
        scanf("%d%d%d", &s, &t, &k);
        dijkstra();
        printf("%d
", astar());
    }
    return 0;
}