bzoj 1598: [Usaco2008 Mar]牛跑步 [k短路 A*] [學習筆記]

1598: [Usaco2008 Mar]牛跑步

題意：k短路

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~~貌似A*的題目除了x數碼就是k短路~~

\[ f(x) = g(x) + h(x) \]
\(g(x)\)為到達當前狀態實際代價，\(h(x)\)為當前狀態到目標狀態的估計代價，需滿足\(h(x) \le 到目標狀態的實際最小代價\)

k短路問題中，\(g(x)\)為當前到x的路徑長度，\(h(x)\)為x到終點的最短路

根據dijkstra演算法，節點i第k次出優先佇列時就是s到i的k短路

但是這個演算法可以被n元環卡成\(O(nk)\)，還有一種做法還要寫可持久化堆好麻煩不學了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define fir first
#define sec second
const int N = 1005, M = 1e4+5, INF = 1e9+5;
inline int read(){
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, k, u, v, w, s, t;
namespace I {
    struct edge{int v, ne, w;} e[M];
    int cnt, h[N];
    inline void ins(int u, int v, int w) { 
        e[++cnt] = (edge){v, h[u], w}; h[u] = cnt;
    }
    priority_queue<pii, vector<pii>, greater<pii> > q;
    int vis[N];
    void dij(int s, int t, int *d) {
        for(int i=1; i<=n; i++) d[i] = INF, vis[i] = 0;
        d[s] = 0; q.push(make_pair(0, s));
        while(!q.empty()) {
            int u = q.top().sec; q.pop(); 
            if(vis[u]) continue; vis[u] = 1;
            for(int i=h[u];i;i=e[i].ne) {
                int v = e[i].v;
                if(d[v] > d[u] + e[i].w) 
                    d[v] = d[u] + e[i].w, q.push(make_pair(d[v], v));
            }
        }
    }
}

int d_t[N];

struct edge{int v, ne, w;} e[M];
int cnt, h[N];
inline void ins(int u, int v, int w) {
    e[++cnt] = (edge){v, h[u], w}; h[u] = cnt;
}
priority_queue<pii, vector<pii>, greater<pii> > q;
int vis[N], ans[105];
void a_star(int s, int t) {
    I::dij(t, s, d_t);
    for(int i=1; i<=n; i++) vis[i] = 0;
    q.push(make_pair(d_t[s], s));
    while(!q.empty()) {
        int u = q.top().sec, d = q.top().fir; q.pop(); 
        vis[u]++;
        if(u == t) { 
            ans[vis[t]] = d;
            if(vis[t] == k) return;
        }
        if(vis[u] <= k) for(int i=h[u];i;i=e[i].ne) {
            int v = e[i].v;
            q.push(make_pair(d - d_t[u] + d_t[v] + e[i].w, v));
        }
    }
}

int main() {
    freopen("in", "r", stdin);
    n=read(); m=read(); k=read(); s=n; t=1;
    for(int i=1; i<=m; i++) u=read(), v=read(), w=read(), I::ins(v, u, w), ins(u, v, w); 
    memset(ans, -1, sizeof(ans));
    a_star(s, t);
    for(int i=1; i<=k; i++) printf("%d\n", ans[i]);
}