POJ_2449_Remmarguts' Date

一直对AStar(简称A*)有些抗拒，感觉是挺难的东西，特别是那个f=g+h的公式里面的参数到底要怎么算总搞不清楚。但是今天把这题给做了，感觉对A*的理解又上一层了。

对于本题,h就是i点到目的地的最短距离，g就是开始地方到i点的实际距离，我们要求f=g+h最小，并且当目的地第K次出队时，那么这时候g的值就是我们所要求解的答案，对于入队大于K次的点，我们不继续求解，因为此后求的就是K+1之后的最短距离了。

怎么求各点到目的点的最短距离呢？只要把所有边反向，然后以目的地为源点求一下最短路，得到的结果就是各点到目的地的最短路。

参考代码：

#include <stdio.h>#include <stdlib.h>#include <string.h>#include <queue>using namespace std;const int MAXN = 2000;const int MAXM = 100100;const int INF = 0x7f7f7f7f;typedef struct Edge {    int to,dis,next;}Edge;typedef struct dNode {    int now;    int dis;    dNode(){};    dNode(int n,int d):now(n),dis(d){};}dNode;typedef struct aNode {    int v,g;    aNode(){};    aNode(int v,int g):v(v),g(g){};}aNode;void init();void add_edge(int form,int to,int dis,Edge * edge,Edge * head,int &total);void dijkstra(int src,Edge * edge,Edge * head);int astar(int src,int des,int K,Edge * edge,Edge * head);bool operator < (const dNode &a,const dNode &b);bool operator < (const aNode &a,const aNode &b);int vis[MAXN];int dis[MAXN];Edge head_1[MAXN];  //记录正向边的邻接表 Edge head_2[MAXN];  //记录反向边的邻接表 Edge edge_1[MAXM];  //记录正向边数据 Edge edge_2[MAXM];  //记录反向边数据 int N,M,total_1,total_2;int main(int argc,char * argv[]) {    int i,j,k;    int from,to,dis;    int S,T,K;    int res;    while (scanf("%d%d",&N,&M)!=EOF) {        init();        for (i=0;i<M;++i) {            scanf("%d%d%d",&from,&to,&dis);            add_edge(from,to,dis,edge_1,head_1,total_1);            add_edge(to,from,dis,edge_2,head_2,total_2);        }        scanf("%d%d%d",&S,&T,&K);        if (S == T) ++K;        dijkstra(T,edge_2,head_2);        res = astar(S,T,K,edge_1,head_1);        printf("%d\n",res);    }    return 0;}void init() {    int i;    for (i=0;i<=N;++i) {        head_1[i].next = 0;        head_2[i].next = 0;    }    total_1 = 0;    total_2 = 0;}void add_edge(int from,int to,int dis,Edge * edge,Edge * head,int &total) {    ++total;    edge[total].to = to;    edge[total].dis = dis;    edge[total].next = head[from].next;    head[from].next = total;}void dijkstra(int src,Edge * edge,Edge * head) {    int p;    dNode node;    priority_queue<dNode> pq;    memset(dis,0x7f,sizeof(dis));    dis[src] = 0;    while (!pq.empty()) pq.pop();    pq.push(dNode(src,0));    memset(vis,0,sizeof(vis));    while (!pq.empty()) {        node = pq.top();        pq.pop();        if (vis[node.now]) continue;        vis[node.now] = 1;        p = head[node.now].next;        while (p) {            if (dis[node.now] < dis[edge[p].to] - edge[p].dis) {                dis[edge[p].to] = dis[node.now] + edge[p].dis;                pq.push(dNode(edge[p].to,dis[edge[p].to]));            }            p = edge[p].next;        }    }}int astar(int src,int des,int K,Edge * edge,Edge * head) {    int p;    aNode now;    priority_queue<aNode> pq;    while (!pq.empty()) pq.pop();    memset(vis,0,sizeof(vis));    pq.push(aNode(src,0));    while (!pq.empty()) {        now = pq.top();        pq.pop();        ++vis[now.v];        if (vis[now.v] > K) continue;        if (K == vis[des]) return now.g;        p = head[now.v].next;        while (p) {            pq.push(aNode(edge[p].to, now.g + edge[p].dis));            p = edge[p].next;        }    }    return -1;}bool operator < (const dNode &a,const dNode &b) {    return a.dis > b.dis;}bool operator < (const aNode &a,const aNode &b) {    return a.g + dis[a.v] > b.g + dis[b.v]; }