[AcWing 1127] 香甜的黄油
作者:互联网
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;
}
- 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