HDU ACM 2066 一个人的旅行

一个人的旅行

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13964    Accepted Submission(s): 4691

 

【题目链接】http://acm.hdu.edu.cn/showproblem.php?pid=2066

【解题思路】题目的意思是说起点有几个，终点也有几个，找起点到终点的映射中最短的那条路，这样的就直接新增两个点，一个是总的起点，车费为0到达题目所述的起点，一个是总的终点，车费为0从题目所述的终点达到这个点，现在问题就转化为求新起点到新终点的最短距离

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

const int NV = 1004;
const int NE = 100004;
const int INF = 1<<30;
int ne, nv, tot, max_node;
bool vis[NV];
int dist[NV];
int aim[NV];
struct Edge{
    int v, cost, next;
    Edge(){}
    Edge(int a, int c){v = a, cost = c;}
    Edge(int a, int c, int d){v = a, cost = c, next = d;}
    bool operator < (const Edge& x) const
    {
        return cost > x.cost;
    }
}edge[NE];
int eh[NV];

void Dij(int s = 0)
{
    for(int i = 1; i <= nv; ++i) dist[i] = i == s ? 0 : INF;
    priority_queue<Edge> que;
    que.push(Edge(s, 0));
    while(!que.empty())
    {
        Edge tmp = que.top();
        int u = tmp.v;
        que.pop();
        if(vis[u]) continue;
        vis[u] = true;
        for(int i = eh[u]; i != -1; i = edge[i].next)
        {
            int v = edge[i].v;
            if(!vis[v] && dist[v] > edge[i].cost + dist[u])
            {
                dist[v] = edge[i].cost + dist[u];
                que.push(Edge(v, dist[v]));

            }
        }
    }
    return;
}

void addedge(int u, int v, int cost)
{
    Edge e = Edge(v, cost, eh[u]);
    edge[tot] = e;
    eh[u] = tot++;
    return;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif
    int s, d;
    while(scanf("%d%d%d", &ne, &s, &d) != EOF)
    {
        memset(eh, -1, sizeof(eh));
        memset(vis, false, sizeof(vis));
        max_node = tot = 0;
        int u, v, cost;
        for(int i = 0; i < ne; ++i)
        {
            scanf("%d%d%d", &u, &v, &cost);
            addedge(u, v, cost);
            addedge(v, u, cost);
            max_node = max_node > u ? (max_node > v ? max_node : v) : (u > v ? u : v);
        }
        for(int i = 0; i < s; ++i)
        {
            int temp;
            scanf("%d", &temp);
            addedge(0, temp, 0);
            addedge(temp, 0, 0);
            if(temp > max_node) max_node = temp;
        }
        max_node++;
        for(int i = 0; i < d; ++i)
        {
            int temp;
            scanf("%d", &temp);
            addedge(max_node, temp, 0);
            addedge(temp, max_node, 0);
            aim[i] = temp;
        }
        nv = max_node+1;
        Dij();
        int ans = INF;
        for(int i = 0; i < d; ++i)
        if(ans > dist[aim[i]]) ans = dist[aim[i]];
        printf("%d
", ans);
    }
    return 0;
}