[SPOJ 30669] Ada and Trip

[题目链接]

https://www.spoj.com/problems/ADATRIP/

[算法]

直接使用dijkstra堆优化算法即可

[代码]

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + ;
const int INF = 2e9;

struct edge
{
        int to,w,nxt;
} e[MAXN << ];

int i,j,tot,a,b,l,n,m,u,q;
int head[MAXN];
pair<int,int> ans;

inline void addedge(int u,int v,int w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
}
inline pair<int,int> dijkstra(int s)
{
        int i,l,r,cur,ans1,ans2,v,w;
        static int dist[MAXN];
        static bool visited[MAXN];
        priority_queue< pair<int,int> > q;
        for (i = ; i < n; i++)
        {
                dist[i] = INF;
                visited[i] = false;
        }
        q.push(make_pair(,s));
        dist[s] = ;
        while (!q.empty())
        {
                cur = q.top().second;
                q.pop();
                if (visited[cur]) continue;
                visited[cur] = true;
                for (i = head[cur]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;
                        if (dist[cur] + w < dist[v])
                        {
                                dist[v] = dist[cur] + w;
                                q.push(make_pair(-dist[v],v));
                        }
                }
        }
        ans1 = ;
        for (i = ; i < n; i++)
        {
                if (dist[i] != INF)
                        ans1 = max(ans1,dist[i]);
        }
        ans2 = ;
        for (i = ; i < n; i++)
        {
                if (dist[i] == ans1)
                        ans2++;
        }
        return make_pair(ans1,ans2);
}

int main()
{

        scanf("%d%d%d",&n,&m,&q);
        for (i = ; i <= m; i++)
        {
                scanf("%d%d%d",&a,&b,&l);
                if (l == ) continue;
                addedge(a,b,l);
                addedge(b,a,l);
        }
        while (q--)
        {
                scanf("%d",&u);
                ans = dijkstra(u);
                printf("%d %d\n",ans.first,ans.second);
        }

        return ;

}