poj 2449 Remmarguts' Date（K短路,A*算法）

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http://poj.org/problem?id=2449

大致题意：给出一个有向图，求从起点到终点的第K短路。

K短路与A*算法详解学长的博客。。。

算法过程

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#define LL long long#define _LL __int64using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 1010;const int maxm = 100010;struct node{int u,v,w;};int s,t,k;int n,m;vector <struct node> edge[maxn],edge1[maxn]; //邻接表存图以及反向图int dis[maxn]; // 终点到所有点的最短路int time[maxn];// 每个点的出队列次数int ans;bool operator > (const struct node &a, const struct node &b){return a.w+dis[a.v] > b.w + dis[b.v];}priority_queue < node, vector<node>, greater<node> >q;void init(){for(int i = 1; i <= n; i++){edge[i].clear();edge1[i].clear();}}//spfa求终点到其他左右点的最短路void spfa() {int inque[maxn];queue<int> que;while(!que.empty()) que.pop();memset(inque,0,sizeof(inque));memset(dis,INF,sizeof(dis));dis[t] = 0;inque[t] = 1;que.push(t);while(!que.empty()){int u = que.front();que.pop();inque[u] = 0;for(int i = 0; i < (int)edge1[u].size(); i++){int v = edge1[u][i].v;int w = edge1[u][i].w;if(dis[v] > dis[u] + w){dis[v] = dis[u] + w;if(!inque[v]){inque[v] = 1;que.push(v);}}}}}void solve(){while(!q.empty()) q.pop();memset(time,0,sizeof(time));struct node tmp;bool flag = false;//起点进队列tmp.v = s;tmp.w = 0;q.push(tmp);while(!q.empty()){struct node u = q.top();q.pop();time[u.v]++;if(time[u.v] >= k) //出队次数大于等于K时{if(u.v == t) //如果是终点，判断与起点是否相同//若不相同,当前值便是第K短路，否则第K+1次才是最短路{if(t != s || (t == s && time[u.v] == k+1)){flag = true;ans = u.w;break;}}if(time[u.v] > k)//如果不是终点，当出队次数大于K次就不再进队列continue;}int now = u.v;for(int i = 0; i < (int)edge[now].size(); i++){struct node tmp;tmp.v = edge[now][i].v;tmp.w = u.w + edge[now][i].w;q.push(tmp);}}if(!flag)ans = -1;}int main(){while(~scanf("%d %d",&n,&m)){init();int u,v,w;for(int i = 1; i <= m; i++){scanf("%d %d %d",&u,&v,&w);edge[u].push_back( (struct node){u,v,w} );edge1[v].push_back( (struct node) {v,u,w} );}scanf("%d %d %d",&s,&t,&k);spfa();solve();printf("%d\n",ans);}return 0;}

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