HDU6166(dijkstra)

最新推荐文章于 2020-02-25 13:36:59 发布

creatorx

最新推荐文章于 2020-02-25 13:36:59 发布

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分类专栏：最短路

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本文介绍了一道复杂的图论问题，目标是最小化选定节点间的最短路径距离。通过巧妙地将节点划分为两个集合并计算它们之间的最小距离，最终解决了SeniorPan面临的难题。

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Senior Pan

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 722 Accepted Submission(s): 289

Problem Description

Senior Pan fails in his discrete math exam again. So he asks Master ZKC to give him graph theory problems everyday.
The task is simple : ZKC will give Pan a directed graph every time, and selects some nodes from that graph, you can calculate the minimum distance of every pair of nodes chosen in these nodes and now ZKC only cares about the minimum among them. That is still too hard for poor Pan, so he asks you for help.

Input

The first line contains one integer T, represents the number of Test Cases.1≤T≤5.Then T Test Cases, for each Test Cases, the first line contains two integers n,m representing the number of nodes and the number of edges.1≤n,m≤100000
Then m lines follow. Each line contains three integers

xi,yi representing an edge, and

vi representing its length.1≤

xi,yi ≤n,1≤

vi ≤100000
Then one line contains one integer K, the number of nodes that Master Dong selects out.1≤K≤n
The following line contains K unique integers

ai , the nodes that Master Dong selects out.1≤

ai ≤n,

ai !=aj

Output

For every Test Case, output one integer: the answer

Sample Input

Sample Output

  
   Case #1: 2

Source

2017 Multi-University Training Contest - Team

解题思路：这题也是巧妙，把这些点分为两部分，然后求两部分之间的距离，我们把按每个节点编号中的二进制的0,1来划分集合，这样就能保证任意两点都被考虑了

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 10;
int inf = 0x3f3f3f3f;
int n, m, k;
vector<pair<int, int> > g[maxn];
int dis[maxn];
vector<int> one;
vector<int> zero;
int a[maxn];
bool flag[maxn];
bool visit[maxn];
struct node{
    int u;
    int d;
    node(int _u = 0, int _d = 0){
        u = _u;
        d = _d;
    }
    bool operator <(const node &res) const {
        return d > res.d;
    }
};
priority_queue<node> q;
int dijkstra(int type)
{
    while(!q.empty()) q.pop();
    memset(dis, inf, sizeof(dis));
    for(int i = 1; i <= k; i++)
    {
        flag[a[i]] = true;
    }
    if(type == 0)
    {
        for(int i = 0; i < zero.size(); i++)
        {
            int v = zero[i];
            dis[v] = 0;
            q.push(node(v, 0));
            flag[v] = false;
        }
    }
    else
    {
        for(int i = 0; i < one.size(); i++)
        {
            int v = one[i];
            dis[v] = 0;
            q.push(node(v, 0));
            flag[v] = false;
        }
    }
    while(!q.empty())
    {
        node nx = q.top();
        q.pop();
        int u = nx.u;
        int d = nx.d;
        if(flag[u]) return d;
        for(int i = 0; i < g[u].size(); i++)
        {
            int v = g[u][i].first;
            int w = g[u][i].second;
            if(dis[u] + w < dis[v])
            {
                dis[v] = dis[u] + w;
                q.push(node(v, dis[v]));
            }
        }

    }
    return inf;
}
void init()
{
    for(int i = 1; i <= n; i++)
    {
        g[i].clear();
    }
}
int main()
{
    int T;
    //freopen("C:\\Users\\creator\\Desktop\\in1.txt","r",stdin) ;
    //freopen("C:\\Users\\creator\\Desktop\\out.txt","w",stdout) ;

    scanf("%d", &T);
    int Case = 1;
    while(T--)
    {
        scanf("%d%d", &n, &m);
        init();
        int u, v, w;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &u, &v, &w);
            g[u].push_back(make_pair(v, w));
        }
        scanf("%d", &k);
        for(int i = 1; i <= k; i++)
        {
            scanf("%d", &a[i]);
        }
        int Min = inf + 1;
        for(int i = 0; i <= 20; i++)
        {
            one.clear();
            zero.clear();
            for(int j = 1; j <= k; j++)
            {
                int num = (a[j]>>i)&1;
                if(num) one.push_back(a[j]);
                else zero.push_back(a[j]);
            }
            Min = min(Min, dijkstra(1));
            Min = min(Min, dijkstra(0));
        }
        if(k == 1) Min = 0;
        printf("Case #%d: %d\n", Case++, Min);
    }
    return 0;
}