其他分享
首页 > 其他分享> > [AcWing 1127] 香甜的黄油

[AcWing 1127] 香甜的黄油

作者:互联网

image
image

选一个起点,到其他点的最短距离之和最小

堆优化 dijkstra (太慢)

复杂度 \(O(n \cdot log(m) \cdot p) = 500 \times log(1450) \times 800 = 1.2 \times 10^7\)


点击查看代码
#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;

int n, m, p;
int id[N];
int h[N], e[N], ne[N], w[N], idx;
int d[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx ++;
}

void dijkstra(int sp)
{
    memset(d, 0x3f, sizeof d);
    memset(st, false, sizeof st);
    priority_queue<PII, vector<PII>, greater<PII>> heap;
    heap.push({0, sp});
    d[sp] = 0;
    while (heap.size()) {
        auto t = heap.top();
        heap.pop();
        auto v = t.second;
        if (st[v])
            continue;
        st[v] = true;
        for (int i = h[v]; i != -1; i = ne[i]) {
            int j = e[i];
            if (d[j] > d[v] + w[i]) {
                d[j] = d[v] + w[i];
                heap.push({d[j], j});
            }
        }
    }
}

void solve()
{
    cin >> n >> p >> m;
    memset(h, -1, sizeof h);
    for (int i = 1; i <= n; i ++)
        cin >> id[i];
    for (int i = 0; i < m; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
        add(b, a, c);
    }
    int res = INF;
    for (int i = 1; i <= p; i ++) {
        dijkstra(i);
        int sum = 0;
        for (int j = 1; j <= n; j ++) {
            int dist = d[id[j]];
            if (dist == INF) {
                sum = INF;
                break;
            }
            sum += dist;
        }
        res = min(res, sum);
    }
    cout << res << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    solve();

    return 0;
}

  1. 堆优化 \(dijsktra\) 翻车,\(SPFA\) 活了过来

标签:idx,int,memset,st,heap,黄油,sizeof,1127,AcWing
来源: https://www.cnblogs.com/wKingYu/p/16564514.html