CSU 2005: Nearest Maintenance Point（Dijkstra + bitset）

Description

A county consists of n cities (labeled 1, 2, …, n) connected by some bidirectional roads. Each road connects a pair of distinct cities. A robot company has built maintenance points in some cities. Now, they need your help to write a program to query the nearest maintenance point to a specific city.

Input

There will be at most 200 test cases. Each case begins with four integers n, m, s, q(2 ≤ n ≤ 104, 1 ≤ m ≤ 5 × 104, 1 ≤ s ≤ min{n, 1000},1 ≤ q ≤ min{n, 1000}), the number of cities, the number of roads, the number of cities which have maintenance points and the number of queries. Then m lines contain the descriptions of roads. Each of them contains three integers ui, vi, wi(1 ≤ ui, vi ≤ n, ui ≠ vi, 1 ≤ wi ≤ 1000), denoting there is a road of length wi which connects city ui and vi. The next line contains s integers, denoting the cities which have maintenance points. Then q lines contain the descriptions of queries. Each of them contains one integer denoting a specific city. The size of the whole input file does not exceed 6MB.

Output

For each query, print the nearest city which has a maintenance point. If there are multiple such cities, print all of them in increasing order separated by a single space. It is guaranteed that there will be at least one such city.

Sample Input

4 4 2 3
1 2 1
2 3 2
2 4 1
4 3 2
1 3
3
2
4

Sample Output

3
1
1 3

题目大意:有n个城市，m条边将一些城市连接起来，现在有s个城市维护点，q次查询，现在对于每一次查询的城市，输出离他最近的城市维护点，如果有多个则按递增顺序打印。
思路: 把s个城市维护点当起点搜一下最短路，其中用bitset来维护其他城市距离维护点最近的那个起点，对于城市x来说，若有多个起点城市到它的最短距离是相等的就用或运算将之存在bitset，否则直接将最短的那一个维护点赋值给x
　　代码上也做了一些注释（如果对于bitset不是很熟悉的可以向大家推荐一篇感觉还不错的博客。http://www.cppblog.com/ylfeng/archive/2010/03/26/110592.html）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<bitset>
#include<vector>
#include<queue>

using namespace std;
const int INF = 1 << 28;
const int maxn = 20005;
struct node {
    int to, cost;
    node() {}
    node(int a, int b) :to(a), cost(b) {}
    bool operator<(const node&a)const {
        return cost > a.cost;
    }
};
vector<node>e[maxn];
bitset<1005>bt[maxn];
int n, m, s, q;
int dis[maxn], cha[maxn], vis[maxn];
void Dijkstra()
{
    priority_queue<node>Q;
    for (int i = 1; i <= n; i++) {
        dis[i] = INF; bt[i].reset();//初始化距离和清空bitset表
        vis[i] = 0;
    }
    for (int i = 1; i <= s; i++) {
        dis[cha[i]] = 0;
        Q.push(node(cha[i], dis[cha[i]]));
        bt[cha[i]][i] = 1;//bt[i][j] = 1 代表第i个城市最近的城市维护点是第j个
    }
    while (!Q.empty()) {
        node t = Q.top(); Q.pop();
        if (vis[t.to])continue;
        vis[t.to] = 1;//对于每一个点我们只需要以他为起点遍历一次就ok，避免大量的重复计算
        for (int i = 0; i < e[t.to].size(); i++) {
            int tmp = e[t.to][i].to;
            int ct = e[t.to][i].cost;
            if (dis[tmp] > dis[t.to] + ct) {
                dis[tmp] = dis[t.to] + ct;
                Q.push(node(tmp, dis[tmp]));
                bt[tmp] = bt[t.to];//如果有比当前更近的城市维护点则将更近的维护点赋值给现在的点
            }
            else if (dis[tmp] == (dis[t.to] + ct))
                bt[tmp] |= bt[t.to];//如果有多个城市维护点的距离相同就取或赋值
        }
    }
}
int main()
{
    while (scanf("%d%d%d%d", &n, &m, &s, &q) != EOF) {
        for (int i = 1; i <= n; i++)e[i].clear();
        for (int a, b, c, i = 0; i < m; i++) {
            scanf("%d%d%d", &a, &b, &c);
            e[a].push_back(node(b, c));
            e[b].push_back(node(a, c));
        }
        for (int i = 1; i <= s; i++)scanf("%d", &cha[i]);
        sort(cha + 1, cha + s + 1);//因为题目要我们如果有多个点满足要求，则递增输出，所以我们可以事先排个序
        Dijkstra();
        int x;
        while (q--) {
            int a[1010], cnt = 0;
            scanf("%d", &x);
            for (int i = 1; i <= s; i++)
                if (bt[x][i])//对于s个城市维护点来说，若bt[x][i]==1则代表第i个城市到x的距离是最近的
                    a[++cnt] = cha[i];
            for (int i = 1; i < cnt; i++)
                printf("%d ", a[i]);//最后按格式输出即可
            printf("%d
", a[cnt]);
        }
    }
    return 0;
}