[CSU 2005 Nearest Maintenance Point Submit Page] Dijkstra

[CSU 2005 Nearest Maintenance Point Submit Page] Dijkstra

分類：Data Structure Shortest Path Dijkstra

1. 題目鏈接

[CSU 2005 Nearest Maintenance Point Submit Page]

2. 題意描述

有一個n 個頂點m 條邊的無向圖。在這n 個點中有s 個頂點是維修站，然後q 次詢問，每次詢問包含一個頂點u ，輸出距離u 最近的加油站。
如果有多個加油站，按照節點編號從小到大輸出。
數據範圍：(2 ≤ n ≤ 104, 1 ≤ m ≤ 5 × 104) ,(1 ≤ s ≤ minn, 1000,1 ≤ q ≤ minn, 1000)

3. 解題思路

真的是一個很水的最短路。省賽上一眼就看出來思路了。但是當時沒有想到用bitset來記憶化，我用的set，MLE有嘗試縮短鏈的長度。導致全場都沒有調出來，gg...
將所有的加油站當成一個源點。跑一次Dijkstra最短路。然後，對於每次查詢，從查詢的頂點開始dfs（用bitset記憶化每個點能夠到達的加油站）。是的、就是這麼簡單。。。嗚嗚嗚~~~

4. 實現代碼

#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;
}