[CSU 2005 Nearest Maintenance Point Submit Page] Dijkstra

分类：Data Structure Shortest Path Dijkstra

1. 题目链接

[CSU 2005 Nearest Maintenance Point Submit Page]

2. 题意描述

有一个n个顶点m条边的无向图。在这n个点中有s个顶点是维修站，然后q次询问，每次询问包含一个顶点u，输出距离u最近的加油站。
如果有多个加油站，按照节点编号从小到大输出。
数据范围：(2 ≤ n ≤ 104, 1 ≤ m ≤ 5 × 104),(1 ≤ s ≤ minn, 1000,1 ≤ q ≤ minn, 1000)

3. 解题思路

真的是一个很水的最短路。省赛上一眼就看出来思路了。但是当时没有想到用bitset来记忆化，我用的set，MLE有尝试缩短链的长度。导致全场都没有调出来，gg...
将所有的加油站当成一个源点。跑一次Dijkstra最短路。然后，对于每次查询，从查询的顶点开始dfs（用bitset记忆化每个点能够到达的加油站）。是的、就是这么简单。。。呜呜呜~~~

4. 实现代码

#include <map>#include <set>#include <queue>#include <ctime>#include <cstdio>#include <bitset>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 10000 + 100;const int MAXE = 50000 + 100;int n, m, sn, qn, sx[MAXN], qx[MAXN];int idx[MAXN];bitset<1001> bs[MAXN];template<class T>struct Dijkstra {    struct Edge {        T w;        int v, nxt;    } E[MAXE << 1];    typedef pair<T, int> PII;    typedef PII QNode;    int head[MAXN], tot;    T d[MAXN], INF;    void init() {        tot = 0;        memset(head, -1, sizeof(head));    }    void add(int u, int v, T w) {        E[tot].v = v;        E[tot].w = w;        E[tot].nxt = head[u];        head[u] = tot++;    }    priority_queue<QNode, vector<QNode>, greater<QNode> > Q;    void run() {        int u;        memset(d, 0x3f, sizeof(d)); INF = d[0];        while(!Q.empty()) Q.pop();        for(int i = 1; i <= sn; ++i) Q.push(PII(d[sx[i]] = 0, sx[i]));        while(!Q.empty()) {            PII ftp = Q.top(); Q.pop();            u = ftp.second;            if(ftp.first != d[u]) continue;            for(int i = head[u]; ~i; i = E[i].nxt) {                int v = E[i].v; T w = E[i].w;                if(d[u] + w < d[v]) {                    d[v] = d[u] + w;                    Q.push(PII(d[v], v));                }            }        }    }    void dfs(int u) {        int v; T w;        if(bs[u].any() != 0) return;        for(int i = head[u]; ~i; i = E[i].nxt) {            v = E[i].v; w = E[i].w;            if(d[v] + w != d[u]) continue;            dfs(v);            bs[u] |= bs[v];        }    }    void solve() {        for(int i = 1; i <= n; ++i) bs[i].reset();        for(int i = 1; i <= sn; ++i) bs[sx[i]].set(idx[sx[i]]);        for(int i = 1; i <= qn; ++i) {            dfs(qx[i]);            for(int j = 1; j <= sn; ++j) {                if(bs[qx[i]][j] == 0) continue;                printf("%d ", sx[j]);            }            printf("\n");        }    }};Dijkstra<int> dij;int main() {#ifdef ___LOCAL_WONZY___    freopen("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    int u, v, w;    while(~scanf("%d %d %d %d", &n, &m, &sn, &qn)) {        dij.init();        for(int i = 1; i <= m; ++i) {            scanf("%d %d %d", &u, &v, &w);            dij.add(u, v, w);            dij.add(v, u, w);        }        for(int i = 1; i <= sn; ++i) scanf("%d", &sx[i]);        sort(sx + 1, sx + sn + 1);        memset(idx, 0, sizeof(idx));        for(int i = 1; i <= sn; ++i) idx[sx[i]] = i;        for(int i = 1; i <= qn; ++i) scanf("%d", &qx[i]);        dij.run();        dij.solve();    }#ifdef ___LOCAL_WONZY___    cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl;#endif // ___LOCAL_WONZY___    return 0;}