POJ 2249 Remmarguts' Date

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求最K短路。

A星＋SPFA。

其实A星算法的思想挺容易理解的。

关键在于优先队列的使用和 SPFA预处理将要走的路径长度。

保证每次选择一条最短路径来走。

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>#include <iostream>#include <cstdlib>using namespace std;const int maxn = 1005;const int maxm = 500010;const int INF = 1000000000;struct Edge{int v, next, w;}edge[maxm], reedge[maxm];int g[maxn], rg[maxn]; int dist[maxn];int vis[maxn];int n, e, m;int src, des, k;queue <int> que;void addedge(int u, int v, int w){e++;edge[e].v = v;edge[e].next = g[u];edge[e].w = w;g[u] = e;reedge[e].v = u;reedge[e].next = rg[v];reedge[e].w = w;rg[v] = e;}void spfa(){int i;for(i = 0; i <= n; i++) dist[i] = INF;while(!que.empty()) que.pop();memset(vis, 0, sizeof(vis));que.push(des);dist[des] = 0;while(!que.empty()){int u = que.front();que.pop();vis[u] = 0;for(i = rg[u]; i; i = reedge[i].next){int v = reedge[i].v;if(dist[v] > reedge[i].w + dist[u]){dist[v] = reedge[i].w + dist[u];if(!vis[v]) que.push(v);vis[v] = 1;}}}}   //从终点到源点构造每一个点到终点的最短径struct A{int f, g, u;bool operator < (const A &tmp) const {if(tmp.f == f)return tmp.g < g;return tmp.f < f;}}; //g为走到u点所经过的距离。//f = g + dist[u] 表示从u点到终点的最短距离。//用优先队列来维护使得每次选择的路径都是当前最短的,直到走到第k短路。//特别当src ＝ des时，由于这条路不能为0，因而k要加1。priority_queue< A> pq;int aStar(){int cnt = 0;if(des == src) k++;if(dist[des] == INF) return -1;A t, tt;t.g = 0;t.f = t.g + dist[src];t.u =des;while(!pq.empty()) pq.pop();pq.push(t);while(!pq.empty()){t = pq.top();pq.pop();if(t.u == src){cnt ++;if(cnt == k) return t.g;}for(int i = g[t.u]; i; i = edge[i].next){int v = edge[i].v;tt.u = v;tt.g = t.g + edge[i].w;tt.f = tt.g + dist[v];pq.push(tt);}}return -1;}void init(){memset(g, 0, sizeof(g));memset(rg, 0, sizeof(rg));e = 0;}int main(){while(~scanf("%d%d",&n, &m)){init();int i;int u, v, w;for(i = 0; i < m; i++){scanf("%d%d%d", &u, &v, &w);addedge(u, v, w);}scanf("%d%d%d", &src, &des, &k);spfa();printf("%d\n", aStar());}return 0;}

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